Solved Problems

(a) The magnetic field inside a solenoid along the central axes is $B={\mu }_0nI$ where $n=N/L$ is the number of turns per unit length of the solenoid.

First, convert the magnetic field from gauss to the Tesla and next find the $n$ \begin{gather*} 2.34\,G \times \left(\rm{\frac{1\, T}{10^{4}\,G}}\right)=2.34\times 10^{-4}\quad {\rm T} \\ \\ n=\frac{N}{L}=\frac{202}{0.926}=218\quad {\rm m^{-1}}\end{gather*} Therefore, using the formula of magnetic field inside the solenoid $B=\mu_0 nI$ and solving for the current , we get \[I=\frac{2.34\times {10}^{-4}}{\left(4\pi \times {10}^{-7}\right)(218)}=0.854\quad {\rm A}\]

(b) The total magnetic flux through a solenoid with $N$ turns is defined by ${\rm{\Phi }}_M=NBA$ so first find the cross-section of the solenoid 
\begin{align*}A&=\pi r^2\\ \\&=\pi \left(\frac d2\right)^2\\ \\&=\frac{\pi}4(0.122)^2\\ \\&=1.17\times 10^{-2}\quad {\rm m^2}\end{align*} 
\begin{align*}\Phi_m&=BA\\&=(2.34\times 10^{-4})(1.7\times 10^{-2})\\ &=2.74\times {10}^{-6}\quad {\rm Wb}\end{align*} 

The electric force defined by $\vec{F}=q\vec{E}$ so \begin{align*} {\vec{F}}_E&=\left(1.6\times {10}^{-19}{\rm C}\right)(1200\hat j)\\ \\&=1.92\times {10}^{-16}\,\hat j \quad {\rm N}\end{align*}
The force that exerts on a moving charged particle with velocity $\vec v$ in a magnetic field $\vec B$ is found by 
\begin{align*} \vec{F}&=q\vec{v}\times \vec{B}\\ \\&=(1.6\times 10^{-19})\left(2\times 10^{6}\,\hat i\right)\times \left(2\times {10}^{-3}\hat{i}+3\times {10}^{-3}\hat{k}\right)\\ \\&=-9.6\times {10}^{-16}\,\hat{j}\quad {\rm N}\end{align*}
The total force is determined by the vector sum of them as \begin{align*}\vec{F}&={\vec{F}}_E+{\vec{F}}_B\\ \\ &=1.92\times {10}^{-16}\,\hat{j}+(-9.6\times {10}^{-16})\,\hat{j}\\ \\&=-7.68\times {10}^{-16}\,\hat{j}\quad {\rm N}\end{align*}
Now write the net force above in the component form and then set $F=0$ to find the desired speed $v$ as follows: \begin{gather*}F=eE_y-ev_xB_z=0\\ \\ \Rightarrow v_x=\frac{E_y}{B_z}=\frac{1200}{0.003}=4\times {10}^5\quad {\rm m/s}\\ \\ \Rightarrow \vec{v}=4\times {10}^5\,(\hat{i}) \quad{\rm m/s}\end{gather*}

Note: a moving point charge $q$ with velocity $\vec{v}$ produces a magnetic field $\vec{B}$ in space that is given by 
\begin{gather*} \vec{B}=\frac{{\mu }_0}{4\pi }\frac{q\vec{v}\times \hat{r}}{r^2}\\ \\ {\mu }_0=4\pi \times {10}^{-7}\frac{\mathrm{T.m}}{\mathrm{A}}\end{gather*} Where $\hat{r}$ is the unit vector that points to the point P. (in this problem $\hat{r}=\hat{j}$)

When the particle is instantaneously at origin, we have $\vec{v}=280\ \hat{i}\ ,\ {\vec{B}}_P=-0.30\,\mu\mathrm{T\ }\hat{k}$ so  
\begin{align*}\vec{B}&=\frac{{\mu }_0}{4\pi }\frac{q\vec{v}\times \hat{r}}{r^2}\\ \\ -0.30\times {10}^{-6}\hat{k}&={10}^{-7}\frac{Q\left(280\hat{i}\times \hat{j}\right)}{{\left(0.070\right)}^2}\\ \\&=5.71\times {10}^{-3}Q\hat{k}\\ \\\Rightarrow Q&=52.5\quad {\rm \mu C}\end{align*}

First by using the Ampere's law find the magnetic field inside the wire as 
$\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}$ where $I_{enc}$ is the current passing through the surface enclosed by the curve $C$.  By definition $I_{enc}=JA=J(\pi r^2)$. Therefore,
\[\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}\Rightarrow B_r\left(2\pi r\right)={\mu }_0\left(\pi r^2\right)J_r\Rightarrow B_r={\mu }_0\frac{r}{2}J_r\] 
Since the current density passing through each surface is constant i.e. $J_r=J_R=I/\pi R^2$ so
\[\Rightarrow \ B_r=\frac{{\mu }_0I}{2\pi R^2}r\ \ ,\ \ r<R\] 
Substitute $r=R$ to find the magnetic field at the surface of the wire i.e. $B_R={\mu }_0I/2\pi R$
\[B_r=0.36B_R\Rightarrow \ \frac{{\mu }_0I}{2\pi R^2}r=0.36\frac{{\mu }_0I}{2\pi R}\mathrm{\Rightarrow }\mathrm{r=0.36\ R}\] 
 

If a charged particle moves with velocity $\vec{v}$ in a magnetic field $\vec{B}$, then from the field exerts a force on it that is determined by $\vec{F}=q\ \vec{v}\times \vec{B}$. Therefor, 
\begin{align*} \vec{F}&=q\ \vec{v}\times \vec{B}\\ &=\left(-5\right)\left(\hat{i}+7\hat{j}\right)\times \left(10\hat{k}\right)\\ &=-50\left(\hat{i}\times \hat{k}\right)-350\left(\hat{j}\times \hat{k}\right)\\ &=-50\ \left(-\hat{j}\right)-350\, \hat{i}\quad {\rm N} \end{align*}

(a) Since the particle moves without deflecting its path, it is in equilibrium. Two forces acting on it. One is the gravity and the other is the magnetic force. But gravity is in the $-y$ direction so the magnetic force must be in the $+y$ direction. Using the right-hand rule, we can infer that the particle must move below the wire.

Right-hand rule: point thumb in direction of the velocity and fingers in the magnetic field direction then your palm direction is the direction of the magnetic force.
\begin{gather*} mg=qvB_{long\ wire}=qv\left(\frac{{\mu }_0I}{2\pi r}\right)\\ \\ \Rightarrow v=\frac{2\pi rmg}{{\mu }_0qI}\end{gather*} Substituting the known values into the above, we get \begin{align*} v&=\frac{2\pi (0.008)(0.003)(9.8)}{\left(4\pi \times {10}^{-7}\right)\left(6\times {10}^{-6}\right)40}\\ \\ &=4.9\times {10}^6 \quad {\rm m/s}\end{align*}
(b) Stated above. 

By definition, the magnetic field of a long straight wire carrying current $I$ at distance $r$ from it is $B=\frac{{\mu }_0I}{2\pi r}$ so we must first find the current that passes through the wire.
\[I\left(t\right)=\frac{\mathrm{\Delta }q}{\mathrm{\Delta }t}=\frac{22}{2\times {10}^{-3}}=1.1\times {10}^2\mathrm{A}\] 
\[B\left(r\right)=\frac{{\mu }_0I}{2\pi r}=4\pi \times {10}^{-7}\frac{1100}{2\pi \times 50}=44\times {10}^{-7}=440\, \mu\mathrm{T}\] 
 

(a) Since the particle moves in a circular orbit so there is a centripetal force that is provided by the magnetic force. By definition, this centripetal force is related to the velocity of the particle via \[F_c=m\frac{v^2}{r}\] 
Now from the kinetic energy calculate the velocity of the charged particle
\begin{align*} K&=\frac{1}{2}mv^2\\ \\ \Rightarrow v&=\sqrt{\frac{2K}{m}}\\ \\&=\sqrt{\frac{2\times 15.2\times {10}^6\times 1.6\times {10}^{-19}}{3.45\times {10}^{-26}}}\\ \\&=1.19\times {10}^7\quad {\rm m/s}\end{align*} Therefore, we have  \begin{gather*} F_c=F_{magnetic}=m\frac{v^2}{r}\\ \\ \Rightarrow q\vec{v}\times \vec{B}=\frac{mv^2}{r}\end{gather*} But the velocity is perpendicular to the magnetic force ($\theta =90{}^\circ $), therefore \begin{align*} qvB&=\frac{mv^2}{r}\\ \\ \Rightarrow q&=\frac{mv}{Br}\\ \\&=\frac{\left(3.45\times {10}^{-26}\right)\left(1.19\times {10}^7\right)}{\left(2.88\times {10}^{-8}\right)\left(9890\times {10}^3\right)}\\ \\&=1.44\times {10}^{-18}\quad \mathrm{C}\end{align*}
(b) From the relation $q=ne$ the above charge equals to the following number of electrons \begin{align*} q&=ne\\ \\ \Rightarrow n&=\frac{q}{e}\\ \\ &=\frac{1.44\times {10}^{-18}}{1.6\times {10}^{-19}}\\ \\&=9\end{align*} That is nine electrons must be added to this ion to make it natural.
 

(a) Consider the desired point to be a distance $x$ from the wire $I_1$(along the line between the two currents ).

So the total magnetic fields due to the currents $I_1$ and $I_2$ are
\begin{gather*} {\vec{B}}_1=\frac{{\mu }_0I_1}{2\pi x}\hat{y} \\ \\ {\vec{B}}_2=\frac{{\mu }_0I_2}{2\pi \left(d-x\right)}\left(-\hat{y}\right)\\ \\\Rightarrow {\vec{B}}_q={\vec{B}}_1+{\vec{B}}_2 \\ \\ \Rightarrow  {\vec{B}}_q=\frac{{\mu }_0}{2\pi }\left(\frac{I_1}{x}-\frac{I_2}{d-x}\right)\hat{y}\end{gather*} We want the ${\vec{B}}_P$ to be zero, so 
\begin{gather*} {\vec{B}}_q=0\\ \\ \Rightarrow I_1\left(d-x\right)\\ \\ =I_2x\\ \\ \to x\left(I_2+I_1\right)=I_1d \\\ \\ \Rightarrow x=\frac{I_1}{I_1+I_2}d\end{gather*} Substituting the values get \begin{align*} x&=\frac{5}{5+15}(0.04)\\\\ &=0.01\quad {\rm m} \\ \\&=1\quad {\rm cm}\end{align*}
Note: the direction of magnetic field of a straight wire is determined by the right hand rule: put the thumb in the direction of current flow, the fingers curl in the direction of the magnetic field.
 

(b) Note: The direction of the magnetic field at a point around a straight wire is tangent to a circle with the center of the current-carrying wire (see figure below). 
\begin{align*} B_1&=\frac{{\mu }_0I_1}{2\pi r_1}\\\\&=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\right)}{2\pi \times 0.03}\\\\&=3.3\times {10}^{-5}\quad \mathrm{T}\\ \\ B_2&=\frac{{\mu }_0I_2}{2\pi r_2}\\ \\&=\frac{\left(4\pi \times {10}^{-7}\right)\left(15\right)}{2\pi \times \sqrt{{\left(0.03\right)}^2+{\left(0.04\right)}^2}}\\ \\&=6\times {10}^{-5}\quad \mathrm{T}\end{align*} From the right hand rule, $B_1$ is in the $x$ direction but $B_2$ makes an angle $\theta $ with the positive $x$ direction (below axes) so must be decomposed into $x$ and $y$ directions \begin{align*} B_{2x}&=B_2{\mathrm{cos} \theta}\\ \\&=\left(6\times {10}^{-5}\right)\left(\frac{0.03}{0.05}\right)\\ \\&=3.6\times {10}^{-5}\quad \mathrm{T}\\ \\ B_{2y}&=B_2{\mathrm{sin} \theta}\\ \\&=-\left(6\times {10}^{-5}\right)\left(\frac{0.04}{0.05}\right)\\ \\&=-4.8\times {10}^{-5}\quad \mathrm{T}\end{align*} Therefore, by applying the superposition principle, we have \begin{align*}{\vec{B}}_P&={\vec{B}}_1+{\vec{B}}_2\\&=\left(3.3+3.6\right)\times {10}^{-5}\hat{x}+4.8\times {10}^{-5}\left(-\hat{y}\right)\\ \\ \left|{\vec{B}}_P\right|&=\sqrt{B^2_{Px}+B^2_{Py}}\\ &=\sqrt{{\left(6.9\times {10}^{-5}\right)}^2+{\left(-4.8\times {10}^{-5}\right)}^2}\\&=8.4\times {10}^{-5}\quad \mathrm{T}\end{align*}
To find the direction of a vector with respect to the horizontal axis, use the equation below \begin{align*} \alpha &={\rm tan^{-1} \left(\frac{x\ comp.}{y\ comp.}\right)}\\ \\ &={\rm tan^{-1} \left(\frac{-4.8\times {10}^{-5}}{6.9\times {10}^{-5}}\right)}\\\\&={\rm -34.82^\circ}\end{align*}
Since $B_x>0\ and\ B_y<0$ so the resultant magnetic field lies in the fourth quadrant.
(c) In this point there is a net magnetic field due to the currents $I_1$ and $I_2$ that relates to the force exerted by these currents on a third current via\begin{align*}\vec{F}_P&=I_3{\vec{L}}_3\times {\vec{B}}_P\\&=(10)(2)\left(8.5\times {10}^{-5}\right)\\&=1.7\times {10}^{-3}\quad {\rm N}\end{align*}

(a) By definition \begin{align*} I=\int{\vec{J}\cdot\hat{n}dA} &=\int^a_0{\frac{2I_0}{\pi a^2}\left(1-{\left(\frac{r}{a}\right)}^2\right)}\underbrace{2\pi rdr}_{dA}\\&=\frac{4I_0}{a^2}{\left(\frac{1}{2}r^2-\frac{r^4}{4a^2}\right)}^a_0=\frac{4I_0}{a^2}\left(\frac{1}{2}a^2-\frac{a^4}{4a^2}\right)=\frac{4I_0}{a^2}\frac{a^2}{4}=I_0\end{align*} In the above, the current density is in the direction of the unit vector normal to the surface area of cylinder. i.e. $\vec{J}.\hat{n}=\left|\vec{J}\right|\left|\hat{n}\right|{\mathrm{cos} 0{}^\circ \ }=\left|\vec{J}\right|$ As shown in the figure below

(b) Ampere's law states that the circulation of the magnetic field around a curve $C$ is related to the current passing through the surface $S$ bounded by $C$ i.e.$\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}$ where $I_{enc}$ is the current through any surface bounded by the curve $C$. 
\begin{align*}I_{enc}&=\int{\vec{J}\cdot\hat{n}dA}\\ \\&=\int^r_0{\frac{2I_0}{\pi a^2}\left(1-{\left(\frac{r}{a}\right)}^2\right)}\underbrace{2\pi rdr}_{dA}\\ \\&=\frac{4I_0}{a^2}\left(\frac{1}{2}r^2-\frac{r^4}{4a^2}\right)\end{align*} and the left side of Ampere's law is \begin{gather*}\oint_C{\vec{B}\cdot d\vec{l}}=B\oint_C{dl}={\mu }_o I_{enc}\\ \\ \Rightarrow B=\frac{{\mu }_0I_{enc}}{2\pi r}\end{gather*} Therefore, for inside the cylinder we have,\begin{align*}B_{in}\left(r<a\right)&=\frac{{\mu }_0}{2\pi r}\frac{4I_0}{a^2}\left(\frac{1}{2}r^2-\frac{r^4}{4a^2}\right)\\ \\&=\frac{{\mu }_0I_0}{{\pi a}^2}\left(r-\frac{r^3}{2a^2}\right)\end{align*} and for outside the magnetic field is \begin{align*}B_{out}\left(r>a\right)&=\frac{{\mu }_0}{2\pi r}I_{tot}\\ \\&=\frac{{\mu }_0}{2\pi r}I_0\end{align*}

(a) Similar to previous problems, first find the electric field at the desired distance due to a positive charge element $dq$, then by integrating determine the total electric field.
\[d\vec{E}=d{\vec{E}}_x+d{\vec{E}}_y=\int{k\frac{dq}{r^2}\hat{r}}=k\int^a_0{\frac{\lambda dy}{x^2+y^2}\left({\mathrm{cos} \theta \ }\hat{x}+{\mathrm{sin} \theta \ }\left(-\hat{y}\right)\right)}\] 
From the geometry ${\mathrm{sin} \theta \ }=\frac{y}{\sqrt{x^2+y^2}}$ and ${\mathrm{cos} \theta \ }=\frac{x}{\sqrt{x^2+y^2}}$ so
\begin{align*} \vec{E} &=k\int^a_0{\frac{\lambda dy}{x^2+y^2}\left(\frac{x}{\sqrt{x^2+y^2}}\hat{x}+\frac{y}{\sqrt{x^2+y^2}}\left(-\hat{y}\right)\right)}\\ &= k\lambda {\left(\frac{xy}{x^2\sqrt{x^2+y^2}}\hat{x}+\frac{-1}{\sqrt{x^2+y^2}}\left(-\hat{y}\right)\right)}^a_0\\&=k\lambda \left(\underbrace{\frac{a}{x\sqrt{x^2+a^2}}}_{E_x}\hat{x}+\underbrace{\left(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{x}\right)}_{E_y}\hat{y}\right) \end{align*}
(b) By definition, $\vec{F}=q\vec{E}$ so 
\[F_x=\left(-q\right)E_x=-qk\lambda \ \frac{a}{x\sqrt{x^2+a^2}}=-\frac{kQq}{x\sqrt{x^2+a^2}}\] 
\[F_y=\left(-q\right)E_y=-qk\lambda \left(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{x}\right)=-\frac{kQq}{a}\left(\frac{1}{\sqrt{x^2+a^2}}-\frac{1}{x}\right)\] 
In above we have used the fact that $\lambda =\frac{charge}{length}=\frac{Q}{a}$
(c) If $x\gg a$ then we have from the approximation below
\[\frac{1}{\sqrt{x^2+a^2}}=\frac{1}{\sqrt{x^2\left(1+\frac{a^2}{x^2}\right)}}=\frac{1}{x}{\left(1+\frac{a^2}{x^2}\right)}^{-\frac{1}{2}}=\frac{1}{x}\left(1-\frac{a^2}{2x^2}\right)\] 
\[F_x=-\frac{kQq}{x\sqrt{x^2+a^2}}=-\frac{kQq}{x}\frac{1}{x}\left(1-\frac{a^2}{2x^2}\right)=-\frac{kQq}{x^2}\left(1-\frac{a^2}{{2x}^2}\right)\] 
\[F_y=-\frac{kQq}{a}\left(\frac{1}{x}\left(1-\frac{a^2}{2x^2}\right)-\frac{1}{x}\right)=+\frac{kqQ}{2x^3}a\] 
 

(a) Right-hand rule for a moving charged particle in a magnetic field:
Point your right four fingers in the direction of the velocity of the particle so that your palm is directed in the magnetic field direction then the thumb shows the force.
Note: The force on a negative charge is in exactly the opposite direction to that of on a positive charge
(b) Recall from the magnetic field problems section that the force exerted on a charged particle with a velocity $v$ in a uniform magnetic field $B$ is
$\vec{F}=q\,\vec{v}\times \vec B=qvB\,{\mathrm{sin} 90\ }=0.2\times 3.5\times 2.5=1.75\ \mathrm{N}$ to the right (right hand rule).
(c) In the circular motion questions section, we find out that when a particle moves around a circle it experiences a centripetal force. In this problem this force provided by the magnetic field.
\begin{gather*} \Sigma F_r=ma_r=\frac{mv^2}{r} \\\\ F_B=F_r \\\\ \Rightarrow qvB=\frac{mv^2}{r} \end{gather*} Solving the above expression for $r$ and substituting the numerical values into it gives \begin{align*} r&=\frac{mv}{qB} \\\\ &=\frac{(0.025)(3.5)}{(0.2)(2.5)} \\\\ &=0.175\quad \rm m \\\\ &=17.5\quad \rm cm\end{align*}

(a) Right hand rule for finding the direction of the magnetic field around a current carrying wire: point your thumb in the direction of the conventional current, then curl your fingers around the wire, they point in the direction of the magnetic field. (figure from Wikipedia)
(b) The magnetic field of a current carrying wire at distance $r$ is given by $B={\mu }_0I/2\pi r$. so
$$B_1=\frac{{\mu }_0}{2\pi }\frac{I_1}{r_1}=2\times {10}^{-7}\left(\frac{3.5}{0.05}\right)=1.4\times {10}^{-5}\mathrm{T} \qquad \text{Up direction}$$
$$B_2=\frac{{\mu }_0}{2\pi }\frac{I_2}{r_2}=2\times {10}^{-7}\left(\frac{4}{0.15}\right)=5.3\times {10}^{-6}\mathrm{T} \qquad \text{Down direction} $$ 
$$\vec{B}={\vec{B}}_1+{\vec{B}}_2=+8.67\times {10}^{-6}\mathrm{T} \qquad \text{Up direction}$$
(c) The force exerted on a line segment $l$ with current $I$ in a magnetic field $B$ is given by 
\[\vec{F}=I\vec{l}\times \vec{B}\] 
\[{\vec{F}}_2=I_2{\vec{B}}_1\times {\vec{l}}_2=I_2l_2B_1{\mathrm{sin} 90{}^\circ \ }=I_2\frac{{\mu }_0}{2\pi }\frac{I_1}{r}l_2=2\times {10}^{-7}\frac{3.5\times 4}{0.2}=1.4\times {10}^{-5}l_2\] 
\[\Rightarrow \frac{F_2}{l_2}=1.4\times {10}^{-5}\mathrm{N/m}\] 
 

Since the plane flies on $xy$ plane, the magnetic field produced by the wire at distance $y$is given by 
\[\vec{B}\left(y\right)=\frac{{\mu }_0I}{2\pi y}\hat{k}\] 
First, consider a conductor bar with width $2w$ that moves in the magnetic field of the wire. Find the force exerted on it and then use the definition of the potential electric field to determine the potential difference between any points.
The force on a charged rod is $\vec{F}=q\vec{v}\times \vec{B}=qv\left(\hat{i}\right)\times B\left(\hat{k}\right)=qvB\left(-\hat{j}\right)$
\[V_{BA}=V_B-V_A=\int^B_A{\vec{E}.d\vec{l}}=\int^B_A{\frac{\vec{F}}{q}.d\vec{l}}=\int^B_A{(\vec{v}\times \vec{B}).d\vec{l}}\] 
\[V_{BA}=\int^{d+2w}_d{v\frac{{\mu }_0I}{2\pi y}\left(-\hat{j}\right).dy\hat{j}}=-\frac{{\mu }_0Iv}{2\pi }{\mathrm{ln} \frac{d+2w}{d}\ }\] 
$V_{CD}$ is determines as above.
\[\vec{v}\times \vec{B}=v\hat{i}\times B\left(\hat{k}\right)=bv(-\hat{j})\] 
\[V_{CD}=\int^B_A{(\vec{v}\times \vec{B}).d\vec{l}}=\int^B_A{bv\left(-\hat{j}\right).dx\ \hat{i}}=0\] 

the potential difference between two points is independent of the path between them (because the electric force is conservative). thus  
\[V_{CA}=V_{CE}+V_{EA}\] 
\[V_{CE}=0\ same\ reason\ as\ above\to V_{CA}=V_{EA}=-\frac{{\mu }_0Iv}{2\pi }{\mathrm{ln} \frac{d+w}{w}\ }\] 
In this part, we have a rode $AE$ with length $w$ instead of $2w$. So we have used the result (a) with substituting w instead of 2w.
 

(a) The magnetic moment of a current carrying loop with area $A$ is defined by $\vec{\mu }=IA\hat{n}$ where $\hat{n}$ is the unit vector perpendicular to the plane of the loop. In this case, the both loops have the same area and currents so
\[{\vec{\mu }}_L=IA\hat{n}=I\pi R^2\hat{z}={\vec{\mu }}_R\] 
$L$ and $R$ stands for the left and right loops.
(b) The $B$-field of a circular loop (magnetic dipole) with radius $R$ at a point $z$ on the axes of dipole is given by the following relation 
\[\vec{B}={\mu }_0\frac{IR^2\hat{n}}{2{\left(L^2+R^2\right)}^{\frac{3}{2}}}\] 
So the magnetic field of the left loop at the location of the right loop is 
\[{\vec{B}}_L={\mu }_0\frac{IR^2\hat{n}}{2{\left(L^2+R^2\right)}^{\frac{3}{2}}}\xrightarrow{L\gg R}\frac{{\mu }_0}{2\pi }\frac{\vec{{\mu }_1}}{L^3}\] 
(c) The total potential energy of this system is : $U=-{\vec{\mu }}_2.{\vec{B}}_1$ or $U=-{\vec{\mu }}_1.{\vec{B}}_2$ So 
$U=-{\vec{\mu }}_R.{\vec{B}}_L=-\frac{{\mu }_0}{2\pi }\frac{{\vec{\mu }}_L.{\vec{\mu }}_R}{L^3}<0$, since ${\vec{\mu }}_1={\vec{\mu }}_2$ therefore$=-\frac{{\mu }_0}{2\pi }\frac{{\left|{\mu }_1\right|}^2}{L^3}<0$ .
To separate the loops one has to do work against the potential energy of the system such that the potential energy goes to zero. Hence the work done is:
\[W=-U=\frac{{\mu }_0}{2\pi }\frac{{\left|{\mu }_1\right|}^2}{L^3}>0\] 
 

(a) The direction of the magnetic field of a straight current carrying wire is determined by the right hand rule. Point your thumb in direction of the current then the direction that the fingers curl, is direction of the magnetic field. 
(b) The current through a surface is defined as the flux of the current density vector $\vec{J}$ that passes through this surface that is $I=\int{\vec{J}.\hat{n}\ dA}=\int{J_0\hat{z}.\hat{z}\ dA}=\pi a^2J_0$, where $\hat{n}$ is the unit vector perpendicular to the surface and $\int{dA}$ is the cross sectional area of the wire.
Note: $=\int{J_0\hat{z}.(\hat{z}dA_1+\hat{r}dA_2)}$ , where $dA_1$ is the bottom and $dA_2$ lateral surface of the cylinder.(in cylindrical coordinate) 
(c) From Ampere's law $\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}$
\[r<R\ ,\ \oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}\to B_{in}\left(2\pi r\right)={\mu }_0\int{J_0\hat{z}.\hat{z}dA}={\mu }_0J_0\pi r^2\] 
\[\therefore \ B_{in}=\frac{{\mu }_0J_0}{2r}\] 
(d) $r>R,\ \ \ \ \oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}\to B_{out}\left(2\pi r\right)={\mu }_0J_0\pi a^2\to B_{out}=\frac{{\mu }_0J_0}{2r}a^2$
Now express the total current through the wire in terms of initial current density $I_{out}=\int{\vec{J}.\hat{n}dA}=J_0\int{\hat{z}.\hat{z}dA_{total}}=J_0\pi a^2$ , where $dA_{total}$ is the total cross section of the wire.
\[so\ ,\ \ J_0a^2=\frac{I}{\pi }\] 
Substituting the $I/\pi $  for $J_0a^2$ in the magnetic field of outside, we obtain
\[B_{out}=\frac{{\mu }_0I}{2\pi r}\] 
 

The magnetic field $d\vec{B}$ produced by a current element $Id\vec{l}$ at distance $r$ from it is given by the Bio-Savart law as
\[d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2}\] 
For example, the magnetic field due to a counter clockwise current loop at the center of the circle is found as
\[\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I}{R^2}\int^{2\pi }_0{\underbrace{\left(Rd\theta \right)\hat{\theta }}_{d\vec{l}}\times \hat{r}}=\frac{{\mu }_0I}{2R}(-\hat{k})\] 
Where we have used the polar coordinate and $\hat{\theta }\times \hat{r}=-\hat{k}$. That is the magnetic field is into the page. 
Now first find the magnetic fields due to each of the wire segments at the point $P$. The resultant magnetic field at point $P$ is the vector sum of these fields. 
The magnetic fields due to the straight segments of the wire at the point $P$ are zero since the current element and radial direction $\hat{r}$ from the $Id\vec{l}$ to the field point $P$ are parallel and thus their cross product is zero. The magnitude of the magnetic fields due to the arc segments are found as follows
\[B_{arc}=\frac{{\mu }_0}{4\pi }\int^{\theta }_0{\frac{IRd\theta }{R^2}}=\frac{{\mu }_0}{4\pi }\frac{I\theta }{R}\] 
Therefore, the magnetic field of the $\pi /2\ $arc at the center of arc i.e. $P$ is $B_{\frac{\pi }{2}}=\frac{{\mu }_0I}{8R}$. Using the right hand rule (put your thumb in direction of the loop, the direction of the rotation of the fingers indicates the direction of the magnetic field at that point), we can see that the direction of the $B_{\frac{\pi }{2}}$ is into the page i.e. ${\vec{B}}_{\frac{\pi }{2}}=\frac{{\mu }_0I}{8R}\left(-\hat{k}\right)$. Similarly, for the $\pi /4\ $ segment we have $B_{\frac{\pi }{4}}=\frac{{\mu }_0I}{16R}\left(+\hat{k}\right)$. Thus, the net magnetic field at the point $P$ is 
\[{\vec{B}}_P={\vec{B}}_{\frac{\pi }{2}}+{\vec{B}}_{\frac{\pi }{4}}=\frac{{\mu }_0I}{8R}\left(-\hat{k}\right)+\frac{{\mu }_0I}{16R}\left(+\hat{k}\right)=\frac{{\mu }_0I}{16R}(-\hat{k})\] 
So the net magnetic field points into the plane of the paper.
 

(a) Ampere's law states that the line integral $\oint_C{\vec{B}.d\vec{l}}$ around a closed curve $C$ (Amperian loop) is equals to the current passes through any surface bounded by $C$ times ${\mu }_0$ i.e. 
\[\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_C\] 
Where $I_c$ is the net current through any surface $S$ bounded by the curve $C$. Therefore, first draw an Amperian loop as shown in the figure, then calculate the line integral around it to find the magnetic field inside the solenoid. If the length of solenoid is long in comparison with the cross sectional diameter then $B_{out}\sim \ 0$. So
\[\oint_C{\vec{B}.d\vec{l}}=\int^b_a{{\vec{B}}_{in}.d\vec{l}}+\underbrace{\int^c_b{{\vec{B}}^{'}.d\vec{l}}}_{0}+\underbrace{\int^d_c{{\vec{B}}_{out}.d\vec{l}}}_{0}+\underbrace{\int^a_d{{\vec{B}}^{'}.d\vec{l}}}_{0}\] 
Along sides $bc$ and $da$ the magnetic field of the solenoid ${\vec{B}}^{'}$ is perpendicular to the path and so $\vec{B^{'}}.d\vec{l}=0$. But along $ab$, $B_{in}$ is parallel to the path. In the length $L$ there are $N$ turns that each of which passes once the current $I_0$at time $t=0$ so the total current through the rectangle is $I_C=NI_0$, therefore the Ampere's law is 
\[\oint_C{\vec{B}.d\vec{l}}=B_{in}L={\mu }_0NI_0\Rightarrow B_{in}={\mu }_0\frac{N}{L}I_0\] 
(b) Using the right hand rule, we obtain the direction of the magnetic field inside the solenoid as shown in the sketch. 
(c) Faraday's law states that a time varying magnetic flux produces an induced emf (voltage) $\mathcal{E}$ and an induced current in any closed loop. In the mathematical form this law is 
\[\mathcal{E}=-\frac{d{\mathrm{\Phi }}_B}{dt}=-\frac{d}{dt}\int{\vec{B}.\hat{n}dA}\] 
Let us imagine that we have an imaginary circular loop of radius $r$ and perpendicular to the solenoid axis. There is a time varying magnetic field $B$ passing normally through the cross sectional area of the loop. So the flux change inside the loop induces a voltage that is related to the magnitude of the induced electric field by $\left|E\right|=\left|-d\mathcal{E}/d\ell \right|$, where $\ell $ is the circumference of the loop. Thus using the Faraday's law, we obtain
\[\left|\mathcal{E}\right|=\left|-\frac{d{\mathrm{\Phi }}_B}{dt}\right|\to E\left(2\pi r\right)=\frac{d}{dt}\left(B_{in}\left(t\right)\pi r^2\right)\] 
\[\left|E_{in}\right|=\frac{1}{2\pi r}\frac{d}{dt}\left({\mu }_0\frac{N}{L}I\left(t\right)\right)\left(\pi r^2\right)=\frac{r}{2}\frac{{\mu }_0N}{L}\frac{I_0}{\tau }\] 
(d) Using the Lenz's law, we can determine the direction of induced current in an imaginary closed loop shown in the figure. Lenz's law: the induced current in a loop is in the direction that the magnetic field produced by it opposes the change in the magnetic flux through the area enclosed by the loop. Since, in this case, we have a time varying current ($I(t)$) so the magnetic field inside the solenoid (or any closed loop) is increasing with time. By Lenz's law, the induced current ($I_{ind}$) must be counter clockwise to create an induced magnetic field in opposite direction of the magnetic field of the solenoid. Since the electric field drives the current, hence the induced electric field $\left|E_{in}\right|$ would be tangential and counter clockwise.
 

(a) Use the right hand rule as follows to find the direction of the magnetic field of the solenoid. Point your right thumb along the $I_{sol}$ and let your fingers curl through the center of the loop then your fingers pointing in the direction of the $B_{sol}$ that in this case is to the left.
(b) Using the Ampere's law we can obtain the magnetic field inside the solenoid as $B_{in}={\mu }_0\frac{N}{L}I_{sol}$. By definition, the magnetic flux passing through any surface $A$ is ${\mathrm{\Phi }}_M=\vec{B}.\hat{n}A$ where $\hat{n}$ is the unit vector normal to the surface. Since the magnetic field outside an ideal solenoid is zero so the area of the loop is not $\pi a^2$ but $\pi R^2_{sol}$. Thus we obtain 
\[{\mathrm{\Phi }}_M=B_{in}\left(\pi R^2_{sol}\right)={\mu }_0\frac{N}{L}I_{sol}\left(\pi R^2_{sol}\right)\] 
(c) Since the $I_{sol}$ is decreasing, the magnetic flux through the loop decreases. By Lenz's law, the direction of the induced current must be such that the magnetic field created by it opposes the decreasing original flux through the loop. So the induced current must be in the same direction as $I_{sol}$. 
 

A particle moves in a circular path experiences a centripetal force. This force provided by the moving of the charged particle in a magnetic field. Applying Newton's second law and Lorentz force formulae for the magnetic field, we obtain 
\[F_r=ma_r=mr{\omega }^2\Rightarrow F_r=mr{\left(\frac{2\pi }{T}\right)}^2\] 
Where $T$ is the period of the motion and is related to the angular velocity by $\omega =2\pi /T$
\[{\vec{F}}_m=q\vec{v}\times \vec{B}\] 
Since the velocity of the particle perpendicular to the external magnetic field, the magnitude of the magnetic force is $F_m=qvB$. The velocity of a particle in a circular path is related to the angular velocity by $v=r\omega $.
\[F_r=F_m\] 
\[qvB=mr{\left(\frac{2\pi }{T}\right)}^2\to q\left(\frac{r2\pi }{T}\right)B=mr{\left(\frac{2\pi }{T}\right)}^2\] 
\[\Rightarrow B=\frac{2\pi }{T}\frac{m}{q}\] 
Note: this particle must be in clockwise direction to the force exerted on it by magnetic field provides the centripetal acceleration. Using the right hand rule, points your right hand fingers in the direction of the velocity of the particle such that your palm to be in direction of the magnetic field, your thumb points to the magnetic field, we find the correct direction of the particle.
 

(a) The magnetic field due to a long, straight wire which carries current $I$at distance $r$ around it form concentric circles which itsmagnitude is given by the Ampere's law as follows
\[B=\frac{{\mu }_0I}{2\pi r}\] 
The direction of the magnetic field is in the plane which is perpendicular to the wire and is tangent to the concentric circles. Point your right thumb along the current, the direction of the curling your fingers around the wire represents the direction of the magnetic field (right hand rule).
The magnetic field at the location of the top wire has two contributions, ${\vec{B}}_1$ and ${\vec{B}}_2$. The directions of  ${\vec{B}}_1$ and ${\vec{B}}_2$ at the top wire are perpendicular to the line from wires $1$ and $2$ to the top wire $3$. 
From the figure, it is clear that the $y$ components cancel each other and the $x$ components add. Therefore, 
\begin{align*} B_{tot}\left(3\right) &=B_{1x}+B_{2x}=\left|{\vec{B}}_1\right|{\mathrm{cos} 30{}^\circ \ }+\left|{\vec{B}}_2\right|{\mathrm{cos} 30{}^\circ \ }=2\left|B_{1,2}\right|{\mathrm{cos} 30{}^\circ \ }\\ &=2\frac{{\mu }_0I}{2\pi R}{\mathrm{cos} 30{}^\circ \ }=\frac{\sqrt{3}}{2}\frac{{\mu }_0I}{\pi R} \end{align*}
In above, we have used the fact that the $\left|{\vec{B}}_1\right|=\left|{\vec{B}}_2\right|$. Thus the total magnetic field at location of top wire due to the bottom wires is in direction of $-\hat{i}$ with the magnitude above. 
(b) The force per unit length exerted on a straight wire carrying current $I$ in presence of a uniform external magnetic field $\vec{B}$ is given by $\vec{F}=i\vec{l}\times \vec{B}$. This force is perpendicular to both $\vec{B}$ and the vector $\vec{l}$, which points in the direction of the current. 
Therefore, the magnetic force that the top wire experiences due to the presence of the bottom wires is calculated as follows
\[{\vec{F}}_3=I{\vec{l}}_3\times {\vec{B}}_{1,2}=I\ L_3\left(-\hat{k}\right)\times B_{1,2}\left(-\hat{i}\right)=IL_3\left|B_{1,2}\right|\underbrace{\left(\hat{k}\times \hat{i}\right)}_{\hat{j}}\] 
\[\Rightarrow \frac{{\vec{F}}_3}{L_3}=\frac{\sqrt{3}}{2}\frac{{\mu }_0I^2}{\pi R}\hat{j}\] 

(c) Similar to the part (a), 
\[{\vec{B}}_2=\frac{{\mu }_0I}{2\pi R}\left(-\hat{j}\right)\] 
\[{\vec{B}}_3=\frac{{\mu }_0I}{2\pi R}\left({\mathrm{cos} 30{}^\circ \ }\left(-\hat{i}\right)+{\mathrm{sin} 30{}^\circ \ }\hat{j}\right)\] 
The total magnetic field at location of wire $1$ is determined by the superposition principle as
\[{\vec{B}}_{tot}(1)={\vec{B}}_2+{\vec{B}}_3\] 
\[=\frac{{\mu }_0I}{2\pi R}\left(-\hat{j}\right)+\frac{{\mu }_0I}{2\pi R}\left({\mathrm{cos} 30{}^\circ \ }\left(-\hat{i}\right)+{\mathrm{sin} 30{}^\circ \ }\hat{j}\right)\] 
\[=\frac{{\mu }_0I}{2\pi R}\left({\mathrm{cos} 30{}^\circ \ }\left(-\hat{i}\right)+\left({\mathrm{sin} 30{}^\circ \ }-1\right)\hat{j}\right)\] 
\[\Rightarrow {\vec{B}}_{tot}(1)=-\frac{{\mu }_0I}{2\pi R}\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)\] 

(d) Using the right hand rule find the directions of the magnetic fields due to the wires at the point $P$, then use the superposition principle to determine the total magnetic field at that point. As shown in the figure, the two contribution $B_1$ and $B_2$ cancel each other, so the only remaining field is $B_3$ due to the wire $3$ which is in the $-\hat{i}$ direction.
\[{\vec{B}}_{tot}\left(P\right)={\vec{B}}_3=\frac{{\mu }_0I}{2\pi h}\left(-\hat{i}\right)\] 
Where $h^2+{\left(\frac{R}{2}\right)}^2=R^2$ (from Pythagoras theorem). 
\[\Rightarrow h=\sqrt{R^2-\left(\frac{R^2}{4}\right)}=\frac{\sqrt{3}}{2}R\] 
\[\therefore {\vec{B}}_{tot}\left(P\right)=\frac{{\mu }_0I}{2\pi \left(\frac{\sqrt{3}}{2}R\right)}=\frac{{\mu }_0I}{\sqrt{3}\pi R}\left(-\hat{i}\right)\] 
(e) A moving charge $q$ with velocity $\vec{v}$ in an external magnetic field $\vec{B}$ experiences the following magnetic force $\vec{F}=q\vec{v}\times \vec{B}$. This force is perpendicular to both velocity and magnetic field and its direction is found by the right hand rule. Therefore, 
\[{\vec{F}}_P=q\vec{v}\times \vec{B}=0\] 
Since we want the force acted on the particle immediately after releasing from rest, so its initial velocity is zero i.e. $\vec{v}=0$.
(f) Now in this part, the initial velocity is $\vec{v}=v_0\hat{j}$, so the initial force acted on the particle immediately after releasing is 
\[\vec{F}=q\vec{v}\times \vec{B}=qv_0\hat{j}\times \left|{\vec{B}}_{tot}\left(P\right)\right|\left(-\hat{i}\right)\] 
\[=-qv_0\left|{\vec{B}}_{tot}\left(P\right)\right|\underbrace{\left(\hat{j}\times \hat{i}\right)}_{-\hat{k}}\] 
By substituting $\left|{\vec{B}}_{tot}\left(P\right)\right|$ from part (d) into above, we get 
\[\Rightarrow {\vec{F}}_P=\frac{{\mu }_0Iqv_0}{\sqrt{3}\pi R}\hat{k}\] 
We can explicitly show the direction of the force above, by using the right hand rule. Point your right hand fingers along the velocity and curl them in direction of the magnetic field. Your thumb shows the direction of the magnetic force acted.
 

A moving charge $q$ with velocity $\vec{v}$ in an external magnetic field $\vec{B}$ experiences a force as $\vec{F}=q\vec{v}\times \vec{B}$. Since $F$ is always perpendicular to the magnetic field $B$, it does not work on the charge and cannot change the magnitude of the velocity of the particle but this force can change the direction of the moving particle. In the special case where the velocity of the charged particle is perpendicular to the magnetic field, the particle moves in a circular path. We know, a particle in a circular orbit experiences centripetal force. In this case, this force is provided by the magnetic force. Use Newton's second law to find the radius of the orbit as follows \[\left\{ \begin{array}{rcl} F_r & = &\frac{mv^2}{r} \\ F_B & = &qvB \end{array}\right.\Rightarrow F_r=F_B\] \[\Rightarrow \frac{mv^2}{r}=qvB\Rightarrow r=\frac{mv}{qB}\] $r$ is the radius of the particle in this configuration.If the fingers of the right hand are in the direction of the velocity $\vec{v}$ so that they can be turned toward the $\vec{B}$, the thumb points in the direction of the force $\vec{F}$ (right-hand rule). Using this rule, at the moment that the particle enters in the non-zero region of $\vec{B}$, the force is in the direction of the $-y$ axis. Therefore, the exit coordinate of the particle is $(0,-2r,0)$.
\[(0,-\frac{2mv}{qB},0)\] 
 

A current loop with the magnetic moment $\vec{\mu }$ in an external field $\vec{B}$ experiences a torque $\vec{\mu }\times \vec{B}$ which is perpendicular to both of them. This torque tends to twist the loop in the same direction of the magnetic field. The magnetic moment defines as
\[\vec{\mu }=NIA\hat{n}\] 
Where $\hat{n}$ is the unit vector normal to the surface encompasses by the current loop and $N$ is the number of the loops. $A$ is also the area of the surface.
The magnetic moment of this loop is in the $-z$ axis direction (curl your right hand fingers around the direction of the current in the loop, your thumb points to the direction of the $\vec{\mu }$ ). Decompose the vector field $\vec{B}$ into the $x$ and $z$ components as 
\[\vec{B}=\left|\vec{B}\right|{\cos 60{}^\circ \ }\hat{x}+\left|\vec{B}\right|{\sin 60{}^\circ \ }\hat{z}\] 
Thus the cross product of $\vec{\mu }$ and $\vec{B}$ becomes 
\begin{align*}\vec{\tau }&=\vec{\mu }\times \vec B\\ &=NIA\left(-\hat{z}\right)\times \left(\left|\vec{B}\right|{\cos 60{}^\circ \ }\hat{x}+\left|\vec{B}\right|{\sin 60{}^\circ \ }\hat{z}\right)\\ &=-NIA\left|\vec{B}\right|{\cos 60{}^\circ \ }\underbrace{\left(\hat{z}\times \hat{x}\right)}_{\hat{y}} \end{align*} 
\[\Rightarrow \ \vec{\tau }=-NIA\left|\vec{B}\right|{\cos 60{}^\circ \ }\hat{y}\] 
Substituting the given data into the above, we obtain
\[\vec{\tau }=-\left(10\right)\left(1\mathrm{A}\right)\left(100\times {10}^{-4}\,{\mathrm{m}}^{\mathrm{2}}\right)\left(2\,\mathrm{T}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{=0.1}\left(\hat{y}\right)\mathrm{N.m}\] 

If a current carrying wire inserted into an external magnetic field, it experiences a magnetic force that is defined by the $\vec{F}=i\vec{l}\times \vec{B}$. 
In this problem, a current rectangular loop is into the magnetic field $B_{ext}$ that is produced by the  current carrying straight wire $I$ so we must find the magnetic force acting on each of sides of the loop. Recall that the magnetic field of a straight current carrying wire at distance $r$ around it is 
\[B=\frac{{\mu }_0I}{2\pi r}=\frac{\left(4\pi \times {10}^{-7}\right)\left(30\right)}{2\pi r}=6\times {10}^{-6}\frac{1}{r}\] 
In this case $r$ is the distance of each sides of loop from the wire.
\[{\vec{F}}_1=i{\vec{l}}_1\times {\vec{B}}_{ext}=10\left(0.1\right)\hat{x}\times \frac{6\times {10}^{-6}}{0.01\ \mathrm{m}}\left(-\hat{z}\right)=-(6\times {10}^{-4})\underbrace{\left(\hat{x}\times \hat{z}\right)}_{-\hat{y}}=(6\times {10}^{-4})\ \mathrm{N\ }\hat{y}\] 
${\vec{F}}_1$ is the force acting on the upper side of the loop by the wire.
\[{\vec{F}}_2=i{\vec{l}}_2\times {\vec{B}}_{ext}=10\left(0.1\right)\left(-\hat{y}\right)\times \frac{6\times {10}^{-6}}{r}\left(-\hat{z}\right)=\frac{6\times {10}^{-6}}{r}\underbrace{\left(\hat{y}\times \hat{z}\right)}_{\hat{x}}=\frac{6\times {10}^{-6}}{r}\mathrm{N\ }\hat{\mathrm{x}}\] 
\[{\vec{F}}_3=i{\vec{l}}_3\times {\vec{B}}_{ext}=10\left(0.1\right)\left(-\hat{x}\right)\times \left(\frac{6\times {10}^{-6}}{10.1\times {10}^{-2}\mathrm{m}}\right)\left(-\hat{z}\right)=(0.6\times {10}^{-4})\underbrace{\left(\hat{x}\times \hat{z}\right)}_{-\hat{y}}=-(0.6\times {10}^{-4})\mathrm{N\ }\hat{\mathrm{y}}\] 
\[{\vec{F}}_4=i{\vec{l}}_4\times {\vec{B}}_{ext}=10\left(0.1\right)\hat{y}\times \left(\frac{6\times {10}^{-6}}{r}\right)\left(-\hat{z}\right)=-\frac{6\times {10}^{-6}}{r}\underbrace{\left(\hat{y}\times \hat{z}\right)}_{+\hat{x}}=-\frac{6\times {10}^{-6}}{r}\mathrm{N\ }\hat{\mathrm{x}}\] 
Use the superposition principle, vector addition of each of the forces, to find the net force acting on the loop. 
\[{\vec{F}}_{net}={\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3+{\vec{F}}_4=\left(6-0.6\right)\times {10}^{-4}\mathrm{N\ }\hat{\mathrm{y}}=5.4\times {10}^{-4}\mathrm{N\ }\hat{\mathrm{y}}\] 
The forces $F_2$ and $F_4$ are equal but in opposite direction so  they cancel each other. 
 

The integral above is similar to the Ampere's law. This law states that the linear integral of magnetic field $\vec{B}$ around a closed curve $C$ is proportional to the current passing through the surface bounded by the curve $C$. In the mathematical language 
\[\oint_C{\vec{B}.d\vec{s}}={\mu }_0I_{enc}\] 
Since there is no current passing through the triangular path shown, the integral above must be equal to zero!
\[I_{enc}=0\Rightarrow \ \oint_C{\vec{B}.d\vec{s}}=0\] 
We can show explicitly the result above. 
The dot product of vertical path and magnetic field is zero, since the angle between them is $90{}^\circ $. For the horizontal path, we have
\[\int^a_0{\vec{B}.d\vec{s}}=B\int^a_0{{\mathrm{cos} 0{}^\circ \ }dx}=Ba\] 
The incline path have two components, the dot product of the its vertical component and $\vec{B}$ is zero, so the integral along its horizontal component is found as
\[\int^0_a{\vec{B}.d\vec{s}}=B\int^0_a{{\mathrm{cos} 180{}^\circ \ }dx}=-Ba\] 
Thus, the linear integral $\oint{\vec{B}.d\vec{s}}$ is sum of the integrals above,
\[\oint{\vec{B}.d\vec{s}}=\int^a_0{\vec{B}.d\vec{x}}+\int^{0,b}_{a,0}{\vec{B}.d\vec{s}}+\int^0_b{\vec{B}.d\vec{y}}\] 
\[=Ba-Ba=0\] 

The changing magnetic field creates an induced emf in the loop. Applying the Kirchhoff's loop rule, the sum of the potential differences around a closed loop must be zero, and considering the induced emf $\mathcal{E}$in the circuit, we get
\[\mathcal{E}=\mathrm{\Delta }V_R+\mathrm{\Delta }V_C\to \ \mathcal{E}=\left(-IR\right)+\underbrace{\left(-\frac{Q}{C}\right)}_{V_0\left(1\ -\ e^{-\frac{t}{\tau }}\right)}\] 
Since the voltage across the capacitor increases with time, $V_b-V_a>0$, the induced current must be in clockwise direction otherwise the capacitor being completely discharged through the resistor. We know that the potential difference across resistor in the direction of current is $-IR$. From the given voltage across the capacitor and using the definitions of the current $I=dQ/dt$ and capacitance $C=Q/V$ one can derive the current in the loop. 
\[I=\frac{dQ}{dt}=\frac{d}{dt}\left(CV\right)=C\frac{dV}{dt}=\frac{CV_0}{\tau }e^{-\frac{t}{\tau }}\] 
By substituting the time constant of the $C-R$ circuit $\tau =CR$ into above, we get
\[I=\frac{V_0}{R}e^{-\frac{t}{\tau }}\] 
Using Faraday's law, the emf around the loop is equal to the time derivative of the flux $\mathcal{E}=-d\mathrm{\Phi }/dt$, we can rearrange the loop rule as follows
\[-\frac{d\mathrm{\Phi }}{dt}=-\left(\frac{V_0}{R}e^{-\frac{t}{\tau }}\right)R-V_0\left(1-e^{-\frac{t}{\tau }}\right)\to A\frac{dB}{dt}=V_0\] 
Since the area of the loop $A=\pi r^2$ is constant, we can factor it out of integral. Consequently, the rate of change of the magnetic field is 
\[\frac{dB}{dt}=\frac{V_0}{\pi r^2}\] 
As similar reasoning above for the direction of the current in the loop, by Lenz's law this induced current produces a downward flux that strengthen the original flux from the external $B$. So the external magnetic field must be decreasing.
 

If electron enters with angle $\theta $ into the magnetic field $B$, it moves around a helical path instead of circular path. In this case, we have two motions, one is parallel to the $\vec{B}$ and the other is perpendicular to $\vec{B}$. In the perpendicular case, the electron experiences a centripetal force that is caused by the magnetic force $F_m$. Using the definition of the force acting on a moving charge in the external magnetic field, ${\vec{F}}_m=q\vec{v}\times \vec{B}$, and Newton's second law, we can obtain the radius of the circular path as follows
\[\left|{\vec{F}}_m\right|=qv_{\bot }B\ \ ,\ \ F_r=\frac{mv^2_{\bot }}{r}\] 
Where $v_{\bot }=v\,{\mathrm{sin} \theta \ }$ is the perpendicular component of the velocity and $\theta $ is the angle between $\vec{v}$ and $\vec{B}$. Therefore,
\[qv_{\bot }B=\frac{mv^2_{\bot }}{r}\Rightarrow r=\frac{mv_{\bot }}{qB}=\frac{\left(9.11\times {10}^{-31}\right)\left(3\times {10}^6\right){\mathrm{sin} 45{}^\circ \ }}{(1.6\times {10}^{-19})(0.28)}=4.314\times {10}^{-5}\mathrm{\ m}\] 
The period of circular motion is defined as 
\[T=\frac{L}{v_{\bot }}=\frac{2\pi r}{v_{\bot }}\] 
$L=2\pi r$ is the circumference of the circular motion. The distance the electron moves along the magnetic field is called the pitch of the helical path found by $p=v_{\parallel }T$. Thus 
\[p=v_{\parallel }T=v_{\parallel }\left(\frac{2\pi r}{v_{\bot }}\right)=\frac{2\pi r\left(v{\mathrm{cos} 45{}^\circ \ }\right)}{v{\mathrm{sin} 45{}^\circ \ }}=2\pi r\,{\mathrm{cot} 45{}^\circ \ }\] 
\[=2\pi \left(4.314\times {10}^{-5}\right)=27.10\times {10}^{-5}\mathrm{m}\] 
 

When a moving charge with velocity $\vec{v}$ enters a region of  both electric and magnetic fields, it experiences a force called Lorentz force which is calculated by $\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})$. Thus
\[\vec{F}=9\left(21\ \hat{k}+\left(3\hat{i}\times 13\hat{j}\right)\right)=9\left(21+39\right)\hat{k}=540\ \hat{k}\] 
Now using Newton's second law, find the acceleration of the charged particle as 
\[\vec{a}=\frac{\vec{F}}{m}=\frac{540}{0.002}=270000\ \hat{k}\frac{\mathrm{m}}{\mathrm{s}}\] 
 

Using the Ampere's law, we can find the magnetic field of a highly symmetric configuration such as this case. In the mathematical form is 
\[\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}\] 
$C$ is the closed curve which the line integration $\oint $ is performed around it. $I_{enc}$ is also the current passing through the surface bounded by this curve. Let $a=4\ \mathrm{mm}$ and $b=30\ \mathrm{mm}$ be the inner and outer radius of the cylinder, respectively. $r$ is the radius of the contour $C$.
\[\oint_C{\vec{B}.d\vec{l}}={\mu }_0I_{enc}\Rightarrow B\underbrace{\oint_C{dl}}_{2\pi r}={\mu }_0I_{enc}\] 
The current through a cross sectional of a wire is defined as the flux of the current density vector $\vec{J}$ through the surface or in mathematical form $J=I/A$. Since the current density is uniform throughout the wire so we have $J_C=J_{cylinder}$. Therefore 
\[\frac{I_{enc}}{\pi r^2-\pi a^2}=\frac{I_{tot}}{\pi b^2-\pi a^2}\Rightarrow I_{enc}=\frac{r^2-a^2}{b^2-a^2}I_{tot}\] 
Substituting this into the Ampere's law, we obtain the magnetic field inside the wire at the arbitrary radius $r$
\[B=\frac{{\mu }_0}{2\pi r}\frac{r^2-a^2}{b^2-a^2}I_{tot}\] 
The given data for this problem are $I_{tot}=3\ \mathrm{A,\ r=12\ mm,\ \ }{\mu}_0=4\pi \times {10}^{-7}$
\[B=\frac{4\pi \times {10}^{-7}}{2\pi \left(12\mathrm{\times }{\mathrm{10}}^{\mathrm{-}\mathrm{3}}\mathrm{m}\right)}\frac{{\left(12\mathrm{mm}\right)}^{\mathrm{2}}-{\left(4\mathrm{mm}\right)}^{\mathrm{2}}}{{\left(30\mathrm{mm}\right)}^{\mathrm{2}}-{\left(4\mathrm{mm}\right)}^{\mathrm{2}}}\left(3\right)=7.23\times {10}^{-6}\] 
 

The magnetic flux through a surface $S$ is defined by 
\[{\mathrm{\Phi }}_M=\int{\vec{B}.\hat{n}dA}\] 
Where $\hat{n}$ is the unit vector normal to the surface $S$ and $dA$ is the area element. Since both $\vec{B}$ and $\vec{A}$ are constant so we can rewrite the above equation as ${\mathrm{\Phi }}_M=\vec{B}.\vec{A}$. therefore, 
\[{\mathrm{\Phi }}_M=\left(4\hat{i}+19\hat{j}+24\hat{k}\right).\left(\pi r^2\right)\hat{k}=\pi r^2\left\{4\underbrace{\left(\hat{i}.\hat{k}\right)}_{0}+19\underbrace{\left(\hat{j}.\hat{k}\right)}_{0}+24\underbrace{\left(\hat{k}.\hat{k}\right)}_{1}\right\}\] 
\[\Rightarrow \ {\mathrm{\Phi }}_M=24\pi r^2\] 
Now we must find the radius of the loop. The circumference of the loop is given $L=10\,\mathrm{m}$, so by using $L=2\pi r$, we can find the radius $r$ as follows
\[r=\frac{L}{2\pi }\] 
Substituting this into the ${\mathrm{\Phi }}_M$, we obtain
\[{\mathrm{\Phi }}_M=24\ \pi {\left(\frac{L}{2\pi }\right)}^2=\frac{6}{\pi }L^2=\frac{6}{\pi }{\left(10\right)}^2=191\ \mathrm{T.}{\mathrm{m}}^{\mathrm{2}}\] 

There is a changing magnetic field and so a changing magnetic flux. According to Faraday's law, this changing flux produces an electric field whose line integral around a closed curve is proportional to the rate of change of the magnetic flux through any surface bounded by this curve. That is 
\[\mathcal{E}=\oint_C{\vec{E}.d\vec{l}}=-\frac{d{\mathrm{\Phi }}_M}{dt}=-\frac{d}{dt}\int{\vec{B}.\hat{n}dA}\] 
We must find the electric field at radius $r=6\ \mathrm{cm}$. so consider a circular disk of radius $r>R$, and calculate the changing flux passing through the surface bounded by it and the line integral of the electric field as follows
\[\oint_C{\vec{E}.d\vec{l}}=-\frac{d{\mathrm{\Phi }}_M}{dt}=-\frac{d}{dt}\left(\vec{B}.\vec{A}\right)\] 
\[E\underbrace{\oint_C{dl}}_{2\pi r}=-A\frac{dB}{dt}=-\pi R^2\underbrace{\frac{d}{dt}\left(4t\right)}_{4}\] 
In above $A=\pi R^2$ is the cross sectional area of the solenoid. The total flux passes through this area so we must not substitute the area of the surface bounded by the curve $C$.The electric field is constant everywhere on the circular disk so we can factor it out of the integral. The line integral $\oint_C{dl}$ is the circumference of the loop. Therefore, 
\[E=-2\frac{R^2}{r}\] 
Recall that the magnitude of the force acted on a point charge by an external electric field $E$ is found by 
\[\left|\vec{F}\right|=\left|q\vec{E}\right|=6\times \left|\frac{\left(-2\right){\left(1\times {10}^{-2}\mathrm{m}\right)}^{\mathrm{2}}}{6\times {10}^{-2}}\right|=0.02\ \mathrm{N}\] 
 

Ampere's law relates the line integral of the magnetic field around some closed curve $C$ to the current passes through the surface bounded by that curve. In this statement there is a flaw! As shown in the figure, there are two surfaces that bounded by the curve. From the plane surface passes current $i_C$ but no current exist for Bulging surface! In other words, there is only an electric flux passes through the Bulging surface. Maxwell recognized this flaw and corrected it as follow in the Ampere's law 
\[\oint_C{B.dl}={\mu }_0\left(I+I_d\right)={\mu }_0I+{\mu }_0{\epsilon }_0\frac{d{\mathrm{\Phi }}_E}{dt}\] 
Where $I_d$ called \textbf{Maxwell's displacement current}. This is the generalized form of Ampere's law. 
Thus by definition the induced displacement current is 
\[I_d={\epsilon }_0\frac{d{\mathrm{\Phi }}_E}{dt}={\epsilon }_0\frac{d}{dt}\left(\vec{E}.\vec{A}\right)\] 
\[={\epsilon }_0A\,{\mathrm{cos} \theta \ }\frac{dE}{dt}=\left(8.85\times {10}^{-12}\frac{\mathrm{F}}{\mathrm{m}}\right)\left(1\ {\mathrm{m}}^{\mathrm{2}}\right)\left(\mathrm{6\times }{\mathrm{10}}^{\mathrm{6}}\frac{\mathrm{V}}{\mathrm{m.s}}\right){\mathrm{cos} 60{}^\circ \ }=26.55\ \mu\mathrm{A}\] 
Note: the product of Farad($F$) and Voltage ($V$) is, by definition of capacitance $Q=CV$, the unit of charge Coulomb ($C$), on the other hand, the charge over time is defined as current which its SI unit is Ampere ($A$). 
 

The magnitude of magnetic force that a charged particle $q$ moving with velocity $\vec{v}$ experiences in a uniform magnetic field $\vec{B}$ is $\left|\vec{F}\right|=qvB{\,\sin \theta \ }$, where $\theta $ is the angle between the velocity and magnetic field. Here $\theta =90{}^\circ $ and $v_e=v_p$ so their magnitudes are the same. The only difference can be arising due to being positive or negative of charges. 

The correct answer is B.
 

because of vanishing net force on the electron so the magnitude of forces must be equal i.e. $F_E=F_B$ and oriented in opposite directions. We know that the direction of the forces applied on a charged moving particle is determined by right hand rule. Point your fingers of right hand along the velocity of particle so that the palm facing the direction of the magnetic field. Curl them toward $\vec{B}$. Your extended thumb points in the direction of force $\vec{F}$. 
The electric field is in the $-y$ direction so the magnetic field must be in $+y$ direction. Using right hand rule, $\vec{B}$ points in the page. 

The correct answer is C.
 

The magnetic torque on a dipole is defined as $\vec{\tau }=\vec{\mu }\times \vec{B}$ or its magnitude is 
\[\left|\vec{\tau }\right|=\mu B{\,\sin \theta \ }\] 
Where $\theta $ is the angle between dipole and magnetic field. 
In (I), $\theta =90{}^\circ \Rightarrow \left|\vec{\tau }\right|=\mu B$
In (II), $\theta =0{}^\circ \Rightarrow \ \left|\vec{\tau }\right|=0$
In (V), $\theta =180{}^\circ \Rightarrow \ \left|\vec{\tau }\right|=-\mu B$
In (III) and (IV) the angles are between the states of (I) and (V) that is $0{}^\circ <\theta <180{}^\circ $ so their magnitude must be where between (I) and (V). therefore, the configuration (I) has the greatest magnitude of torque. 

The correct answer is A.
 

A wire of length $l$ carrying current $I$ can experience a force on itself  due to the presence of an external magnetic field. This force, which can be used to define the SI unit amperes, is found by $\vec{F}=i\vec{l}\times \vec{B}$ or its magnitude is $\left|\vec{F}\right|=ilB{\,\sin \theta \ }$, where $\theta $ is the angle between the direction of the current in the wire and magnetic field.
When there is two wires, one of them acts as a source of external magnetic field for the second wire. Recall that the magnetic field of a straight current carrying wire by amperes law is found as 
\[B={\mu }_0\frac{I}{2\pi r}\] 

Therefore, as shown in the figure, the magnetic force on wire $2$ due to the magnetic field of wire $1$ at the location of wire $1$ is calculated as follows
\[F_2=i_2l_2B_1=i_2l_2\left(\frac{{\mu }_0i_1}{2\pi d}\right)\] 
Where $d$ is the distance between the wires. So the force per unit length of wire 2 is 
\[\frac{F_2}{l_2}={\mu }_0\frac{i_1i_2}{2\pi d}=\left(4\pi \times {10}^{-7}\right)\frac{2\times 4}{2\pi (4\times {10}^{-2})}=4\times {10}^{-5}\frac{\mathrm{N}}{\mathrm{m}}\] 
To find the direction of the force use the right hand rule. Point your fingers of right hand along the direction of the current in the wire with the palm facing $\vec{B}$ and wrap your fingers toward the $\vec{B}$. The direction of the upright thumb is the direction of the magnetic force. As shown in the figure, the forces are in the opposite direction so are repulsive. 

The correct answer is C.
 

The direction of a current-carrying wire is found by the right hand rule as point the thumb of your right hand along the direction of the current. Curling the four fingers shows the direction of the magnetic field. In this case four fingers curls toward the top of the page into the solenoid. 

The correct answer is C.
 

Lenz's law says that the induced current in a loop must be in direction that the associated magnetic field produced opposes the change in the magnetic flux through the area enclosed by the loop. 
A decreasing magnetic field has a decreasing flux through the loop. A clockwise current in the loop create a magnetic field into the page so that the corresponding flux compensate the decreasing original flux through the loop. 

The correct answer is B.
 

Faraday's law tells us that a changing magnetic field with time induces an emf and a current in the loop that is $\mathcal{E}=-d{\rm{\Phi}}_B/dt$, where the minus sign is an indication of Lenz's law. 

This induced current implies the presence of an induced electric field $\vec{E}$ that is tangent to the loop. Recall that the line integral of $\vec{E}\cdot d\vec{s}$ over a specific path defines the potential electric. 

In this case, there is a closed loop and an induced electric field so the line integral of the above quantity gives the induced emf in the loop i.e. $\mathcal{E}=\oint{\vec{E}\cdot d\vec{s}}$. 

By equating these relations, we get an alternative and useful form of Faraday's law as \[\oint{\vec{E}\cdot d\vec{s}}=-\frac{d{\mathrm{\Phi }}_B}{dt}\] Therefore, it is sufficient to evaluate both sides of above relation at the location of $R/2$. 
\begin{align*} -\frac{d{\rm{\Phi}}_B}{dt}&=\oint{\vec{E}\cdot d\vec{s}}\\\\ -\frac{d}{dt}\left(BA\right)&=E\oint{ds}\\\\ -\frac{dB}{dt}(\pi r^2)&=E\left(2\pi r\right) \\\\ \Rightarrow \left|\frac{dB}{dt}\right|&=\frac{2E}{r}=\frac{2E}{\frac{R}{2}}\\\\&=4\frac{\left(4.5\times {10}^{-3}\right)}{3\times {10}^{-2}}=0.6\quad {\rm \frac Tm}\end{align*}
In the second equality, we have used the formula of magnetic flux. In the third equality since the area of the cylinder of radius, $R/2$ is constant so it is factored out of the derivative.

The correct answer is C.

The rate at which energy passes through a unit area by electromagnetic waves is described by a quantity called Poynting vector $\vec{S}$ which is defined as 
\[\vec{S}=\frac{1}{{\mu }_0}\vec{E}\times \vec{B}\] 
The magnitude of Poynting vector, by definition, represents power per unit area. Because the fields $E$ and $B$ of a sinusoidal electromagnetic wave vary with time it is great important to evaluate the time average of $S$ over one or more cycles. Hence, the average of $S$, which called intensity of the waves $I$, is 
\[I=S_{ave}=\frac{E^2_{max}}{2{\mu }_0c}=\frac{cB^2_{max}}{2{\mu }_0}\] 
Here the average of Poynting vector is 
\[S_{ave}=\frac{{\left(100\right)}^2}{2\left(4\pi \times {10}^{-7}\right)\left(3\times {10}^8\right)}=13.2\ \mathrm{W}\] 
As mentioned, $S_{ave}=P_{ave}/A$ and $P_{ave}=E_{ave}/t$ where $P_{ave}$ is the power of the source and $E_{ave}$ is the corresponding energy . therefore
\[S_{ave}=\frac{E_{ave}}{At}\] 
\[\Rightarrow E_{ave}=\left(13.2\mathrm{\ W}\right)\left(1\times {10}^{-4}{\mathrm{m}}^{\mathrm{2}}\right)\left(10s\right)=13.2\times {10}^{-3}\mathrm{J=13.2\ mJ}\] 

The correct answer is B.