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Magnetic fields Questions

Questions

A $202$ turns of wire exist on a $92.6\, {\rm cm}$ long solenoid. This solenoid produces a magnetic field of $2.34$ gauss at its center. 
(a) What is the current flowing through the coiled wire of this solenoid?
(b) If this solenoid has a diameter of $12.2\mathrm{\ cm}$ and the B-field is uniform inside this solenoid, what is the magnetic flux through one of the turns inside this solenoid?

(a) The magnetic field inside of 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 find the $n$
\[2.34\ \mathrm{G\,\times }\left(\frac{\mathrm{1T}}{{\mathrm{10}}^{\mathrm{4}}\mathrm{G}}\right)=2.34\times {10}^{-4}\, \mathrm{T}\] 
\[n=\frac{N}{L}=\frac{202}{0.926}=218\ {\mathrm{m}}^{\mathrm{-}\mathrm{1}}\] 
Therefore, $B={\mu }_0nI\Rightarrow I=\frac{2.34\times {10}^{-4}}{\left(4\pi \times {10}^{-7}\right)\left(218\right)}=0.854\ \mathrm{A}$
(b) The total magnetic flux through a solenoid with $N$ turns is defined by ${\mathrm{\Phi }}_M=NBA$ so first find the cross section of the solenoid 
\[A=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2=\frac{\pi }{4}{\left(0.122\right)}^2=1.17\times {10}^{-2}{\mathrm{m}}^{\mathrm{2}}\] 
\[{\mathrm{\phi }}_M=BA=\left(2.34\times {10}^{-4}\right)\left(1.17\times {10}^{-2}\right)=2.74\times {10}^{-6}\mathrm{Wb}\] 
 

A proton ($q=+e$) moves in the positive $x$ direction, in a region with an electromagnetic field. Its velocity is $\vec{v}=\left(2\times {10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}$. The fields are $\vec{E}=\left(1200\frac{\mathrm{N}}{\mathrm{C}}\right)\hat{j}$ and $\vec{B}=\left(2\mathrm{mT}\right)\hat{i}+\left(3\mathrm{mT}\right)\hat{k}$. Find the magnitude and direction of the electric, magnetic and total forces acting on the proton. What speed $v$ would result in a zero total force?

A point charge $Q$ moves on the $x$ axis in the positive direction with a speed of $280\ \mathrm{m/s}$. A point P is on the $y$ axis at $y=+70\, \mathrm{mm}$. The magnetic field produced at the point P, as the charge moves through the origin, is equal to $-0.30\, \mu{\rm T\ }\hat{\mathrm{k}}$. What is the charge $Q$?

A long, straight wire with $3.0\ \mathrm{A}$ current flowing through it produces magnetic field strength $1.0\ \mathrm{T}$ at its surface. If the wire has a radius $R$, where within the wire is the field strength equal to $36\%$ of the field strength at the surface of the wire? Assume that the current density is uniform throughout the wire.

A particle with charge $-5.00\ \mathrm{C}$ initially moves at $\vec{v}=\left(1.00\ \hat{i}+7.00\hat{j}\right)\left(\frac{m}{s}\right)$. If it encounters a magnetic field $\vec{B}=10.00\ \mathrm{T\ }\hat{\mathrm{k}}$,find the magnetic force vector on the particle?

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}\ \mathrm{N}
\end{align*}

A horizontal wire has a $40\ \mathrm{A}$ current moving through it to the left. A $3.00\ \mathrm{g}$ particle, with a charge of $+6.00\ \mu\mathrm{C}$ is moving in a straight line horizontally in the direction of the current. Note that the particle's trajectory is also horizontal, and therefor parallel to the wire, it maintains a constant distance of $8.00\ \mathrm{mm}$ from the wire.
(a) How fast is the particle moving?
(b) Is the particle above or below the wire?

Assume that a lightning bolt can be represented a by long straight line of current. If $22.0\ \mathrm{C}$ of charge passes by in a time of $2.00\times {10}^{-3}$ seconds, what is the magnitude of the magnetic field at a distance of $50.0$ meters from the bolt?

A $15.2\ \mathrm{MeV}$ cosmic ray positively charged ion has a mass of $3.45\times {10}^{-26}\, \mathrm{kg}$ moves in a circular orbit whose orbital plane is perpendicular to a uniform $2.88\times {10}^{-8}\mathrm{T}$ magnetic field. 
(a) If the radius of this circular orbit is $9890\ \mathrm{km}$, what is the charge on this cosmic ray ion?
(b) Approximately how many electrons would this ion be deficit, that is how many extra electrons would this ion need to make it a natural particle?

The following diagrams shows a pair of parallel wires carrying currents $I_1$ and $I_2$. $I_1=5.0\ A$ and $I_2=15.0\ \mathrm{A}$, $d=4.0\ \mathrm{cm}$
(a) At what point on a line between the two wires is the $\vec{B}$ field zero?
(b) What is the magnitude and direction of the $\vec{B}$ field at the point P located $3.0\ \mathrm{cm}$ directly below $I_1$?
(c) At the point P a third wire is placed carrying current $I_3=10\ \mathrm{A}$ parallel to both $I_1$ and $I_2$ but directed into the plane of the page from the edge view. What is the magnitude of the force this $2.0\ \mathrm{m}$ length third wire feels?

(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
\[\left\{ \begin{array}{rcl}
{\vec{B}}_1 & = & \frac{{\mu }_0I_1}{2\pi x}\hat{y} \\ 
{\vec{B}}_2 & = & \frac{{\mu }_0I_2}{2\pi \left(d-x\right)}\left(-\hat{y}\right) \end{array}
\right.\Rightarrow {\vec{B}}_q={\vec{B}}_1+{\vec{B}}_2\] 
\[\Rightarrow \ {\vec{B}}_q=\frac{{\mu }_0}{2\pi }\left(\frac{I_1}{x}-\frac{I_2}{d-x}\right)\hat{y}\] 
We want the ${\vec{B}}_P$ to be zero, so 
\[{\vec{B}}_q=0\Rightarrow I_1\left(d-x\right)=I_2x\to x\left(I_2+I_1\right)=I_1d\Rightarrow x=\frac{I_1}{I_1+I_2}d\] 
\[x=\frac{5}{5+15}\left(0.04\right)=0.01\mathrm{m=1cm}\] 
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 magnetic field at a point around a straight wire is tangent to a circle with the center of current-carrying wire (see figure below). 
\[B_1=\frac{{\mu }_0I_1}{2\pi r_1}=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\right)}{2\pi \times 0.03}=3.3\times {10}^{-5}\mathrm{T}\] 
\[B_2=\frac{{\mu }_0I_2}{2\pi r_2}=\frac{\left(4\pi \times {10}^{-7}\right)\left(15\right)}{2\pi \times \sqrt{{\left(0.03\right)}^2+{\left(0.04\right)}^2}}=6\times {10}^{-5}\mathrm{T}\] 
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
\[B_{2x}=B_2{\mathrm{cos} \theta \ }=\left(6\times {10}^{-5}\right)\left(\frac{0.03}{0.05}\right)=3.6\times {10}^{-5}\ \mathrm{T}\] 
\[B_{2y}=B_2{\mathrm{sin} \theta \ }=-\left(6\times {10}^{-5}\right)\left(\frac{0.04}{0.05}\right)=-4.8\times {10}^{-5}\ \mathrm{T}\] 
Therefore 
\[{\vec{B}}_P={\vec{B}}_1+{\vec{B}}_2=\left(3.3+3.6\right)\times {10}^{-5}\hat{x}+4.8\times {10}^{-5}\left(-\hat{y}\right)\] 
\[\left|{\vec{B}}_P\right|=\sqrt{B^2_{Px}+B^2_{Py}}=\sqrt{{\left(6.9\times {10}^{-5}\right)}^2+{\left(-4.8\times {10}^{-5}\right)}^2}=8.4\times {10}^{-5}\ \mathrm{T}\] 
To find the direction of a vector with respect to horizontal axis, use the equation below
\[\alpha ={tan^{-1} \left(\frac{\mathrm{x\ comp.}}{\mathrm{y\ comp.}}\right)\ }={{{tan}}^{-1} \left(\frac{-4.8\times {10}^{-5}}{6.9\times {10}^{-5}}\right)\ }=-34.82{}^\circ \] 
Since $B_x>0\ and\ B_y<0$ so the resultant magnetic field lie 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
\[{\vec{F}}_P=I_3{\vec{L}}_3\times {\vec{B}}_P=\left(10\right)\left(2\right)\left(8.5\times {10}^{-5}\right)=1.7\times {10}^{-3}\ \mathrm{N}\] 


 

A long, straight, solid cylinder, oriented with its axis in the $z$ direction, carries a current whose current density is $\vec{J}$. The current, though symmetric about $z$ varies as $\vec{J}=\frac{2I_0}{\pi a^2}\left(1-{\left(\frac{r}{a}\right)}^2\right)$ for $r<R$ and $\vec{J}=0$ for $r>R$, where $a$ is the radius of the cylinder, $r$ is the radial distance from the cylinder axis and $I_0$ is the total charge.
(a) Now show that $I_0$ is the total current passing through the entire cross section of the wire.
(b) Using Ampere's law, derive an expression for the magnitude of the magnetic field $\vec{B}$ in the regions $r>a$ and $r<a$

Positive charge $Q$ is distributed along the positive $y$ axis between $y=0$ and $y=a$. A negative point charge $-q$ lies on the positive $x$ axis, a distance $x$ from the origin as shown in the figure. 
(a) Calculate the $x$ and $y$ components of the electric field $E_x$ and $E_y$ produced by the charge distribution $Q$ at points on the positive $x$ axis.
(b) Calculate the $x$ and $y$ components of the force that the charge distribution $Q$ exerts on $q$.
(c) When $x\gg a$ derive simplified expressions for both $F_x$ and $F_y$. Hint: use the approximation that ${\left(1+x\right)}^n=1+nx$ when $x\ll 1$. 
Hint: $\int{\frac{du}{{\left(u^2+a^2\right)}^{\frac{3}{2}}}}=\frac{u}{a^2\sqrt{u^2+a^2}}$ and $\int{\frac{u}{{\left(u^2+a^2\right)}^{\frac{3}{2}}}du\ }=-\frac{1}{\sqrt{u^2+a^2}}$

A charge of $+0.2\ \mathrm{C}$ and mass $0.025\ \mathrm{kg}$ is moving as shown in the magnetic field of $2.5\ \mathrm{T}$ directed out of the page. ($v=3.5\ \mathrm{m/s}$)
(a) In the figure, sketch the trajectory of the charge.
(b) What is the magnitude force exerted on the charge( magnitude and direction)
(c) What would the radius of the trajectory be?

The figure shows two parallel wires carrying currents $I_13.5\mathrm{A}$ and $I_2=4\mathrm{A}$, both out of the page, separated by a distance of $20\ \mathrm{cm}$. 
(a) Sketch the direction of the magnetic fields from wires $1$ and $2$ at point $P$.
(b) Determine the net B field at a point (P) $5\, \mathrm{cm}$ right of wire $1$.
(c) Determine the magnitude and direction of the forced exerted on wire 2 from wire 1.

An airplane with a speed $v$ is flying at the same altitude of a high voltage cable carrying a current $I$ as shown in the figure. Determine the following potential differences :$V_{BA}\ ,\ V_{CD}\ ,\ V_{CA}$. (notation: $V_{ij}=V_i-V_j$) 

Two current loops with equal currents $I$ (both oriented in the same direction) and radii $R$ separated by a long distance $L$ as shown in the figure. $L$ is much larger than $R$.
(a) Find the magnetic moments of both current loops.
(b) Find the magnetic field generated by the loop on the left at the center of the other loop on the right.
(c) Show that, in order to separate these loops infinitely far apart, one has to do a positive work given by 
\[W={\mu }_0\frac{{\vec{\mu }}_1.{\vec{\mu }}_2}{2\pi {\left(L^2+R^2\right)}^{\frac{3}{2}}}\] 

Consider an infinite cylindrical wire of radius $a$ centered along the $z$ axis with current density $\vec{J}=\hat{z}J_0$, where $J_0$ is a constant which is known.
(a) Draw the direction of the magnetic field vector inside and outside of the wire. (No calculation is required for this part)
(b) Determine the total current that passes through this wire.
(c) What is the strength of the magnetic field inside the wire.(as a function of the radial distance $r$ from the wire axis)
(d) What is the strength of the magnetic field outside the wire?.(as a function of the radial distance $r$ from the wire axis)

A current $I_0$ is present in the two circularly arced wire segments of radius $R$ depicted below as thick black lines. One are subtends an angle of $\pi /2$, and the other subtends an angle of $\pi /4$. The lead wires point radially outward from point $P$, the location of the center of the two arced wire segments. Derive from the Bio-Savart law the magnitude and direction of the magnetic field at the point $P$. 

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 solenoid with $N$ turns, length $l$ and radius $R$ has a current $I=I_0(1+t/\tau )$, where $\tau $ and $I_0$ are both positive constants and $t$ is time. 
(a) Derive an expression for the magnetic field inside the solenoid at a radius of $r$ where $r<R$ at time $t=0$.
(b) Draw the direction of the magnetic field inside the solenoid at time $t=0$ on the diagram above which depicts a cross sectional view of the solenoid.
(c) What is the magnitude of the electric field at radius $r<R$ inside the same solenoid at some arbitrary time $t>0$?
(d) Draw the direction of the electric field at point $P$ (at radius $r$) on the diagram below depicting a bottom view of a solenoid with the current moving clockwise. 

A solenoid has $N$ turns, a radius $R_{sol}$, a total length $L$ and a current $I_{sol}$. The solenoid depicted below is a cross sectional view.
(a) Draw the direction of the magnetic field inside the solenoid.
(b) What is the magnetic flux produced by the solenoid on a concentric circular loop of radius $a$?
(c) If $I_{sol}$ is decreasing, what is the direction of the induced current in the loop?

A particle with charge $q$ and mass $m$ is moving in a circular path perpendicular to a constant magnetic field. The particle takes a time $T$ to complete one revolution. Determine the magnitude of the magnetic field in term of $q,m$ and $T$.

Three very long current carrying wires are arranged to form an equilateral triangle with side $R$, as shown in the figure below. The bottom two wires carry current $I$ out of the page, and the top wire carries current $I$ into the page. 
(a) Determine the magnetic field at the location of top wire due to the bottom two wires.
(b) Determine the force per unit length on the top wire due to the bottom wire.
(c) Determine the magnetic field at the location of the lower left wire due to the other two.
(d) Determine the magnetic field at the point $P$ midway between the lower two wires.
(e) A particle with charge $+q$ is released from rest at point $P$. Determine the force on this particle due to the magnetic field immediately after it is released. 
(f) A particle with charge $+q$ is released with initial velocity in the $x$ direction, $\vec{v}=v_0\hat{j}$. Determine the force on this particle due to the magnetic field immediately after it is released. 

The region of space $x>0$ is permeated by a uniform magnetic field $\vec{B}=B\hat{k}$pointing in the $z$ direction; the field is zero in the region $x\ <\ 0$. A particle of charge $q$ and velocity $\vec{v}=v\,\hat{i}$ enters the region of non-zero field at the point with $(x,\ y,\ z)$ coordinates $(0,\ 0,\ 0)$. What are the coordinates of the point where the particle exits the region of nonzero field?

The figure shows a square $\mathrm{(10\ cm\ \times \ 10\ cm)}$, $10$-turn coil of wire, which carries a current of $1.0\ A$. The coil is hinged along one side, as shown in the figure, and is mounted in the $x\ -\ y$ plane, at $60{}^\circ $ to the direction of a uniform magnetic field of magnitude $\mathrm{2.0\ T}$. What is the magnitude of the torque acting on the coil (about the hinge line)?

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
\[\vec{\tau }=-\left(10\right)\left(1\mathrm{A}\right)\left(100\times {10}^{-4}\,{\mathrm{m}}^{\mathrm{2}}\right)\left(2\,\mathrm{T}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{=0.1}\left(\hat{y}\right)\mathrm{N.m}\] 

As shown in the figure, a long straight wire carries a current of $\mathrm{30\ A}$ and is co-planar with a square loop of side $\mathrm{10\ cm}$ which carries a current of $\mathrm{10\ A}$. If the distance between the wire and the nearest side of the loop is $\mathrm{1.0\ cm}$, what is the magnitude and direction of the net force on the loop? Recall that ${\mu }_0=4\pi \times {10}^{-7}\mathrm{T.m/A}$

A uniform magnetic field $\vec{B}=B_0\hat{i}$ points in the $x$-direction. What is the value of the integral $\oint{\vec{B}.d\vec{s}}$ around the closed triangular path shown in the figure? 

A circular-shaped circuit of radius $r$, containing a resistance $R$ and capacitance $C$, is situated with its plane perpendicular to the spatially uniform magnetic field within which it is immersed. The magnetic field is directed into the page as shown. Starting at time $t\ =\ 0$, the voltage difference $V_{ba}=V_b\ -V_a$ across the capacitor plates is observed to increase with time according to $V_{ba}=V_0\left(1\ -\ e^{-\frac{t}{\tau }}\right)$ where $V_0$ and $\tau $ are positive constants. Determine $dB/dt$, the rate at which the magnetic field magnitude changes with time. Is $B$ becoming larger or smaller as time increases?

An electron enters a uniform magnetic field at a $45{}^\circ $ angle to the magnetic field $B$. Determine the radius $r$ and pitch $p$ (distance between loops) of the electron's helical path assuming its speed is $3\times {10}^6\mathrm{m/s}$.

A particle with charge $9.0\ \mathrm{C}$ moving at $\vec{v}=3\ \hat{i}\, \mathrm{m/s}$ enters a magnetic field $\vec{B}=13\ \hat{j}\mathrm{m/s}$ and an electric field $\vec{E}=21\ \hat{\mathrm{k}}\, \mathrm{N/C}$. find the acceleration of the particle it its mas is $0.00200\ \mathrm{kg}$.

A hollow cylinder with an inner radius of $\mathrm{4.0\ mm}$ and an outer radius of $\mathrm{30\ mm}$ conducts a $\mathrm{3.0\ A}$ current. What is the magnitude of the magnetic field at a point $\mathrm{12\ mm}$ from its center? Assume that the current density is uniform throughout the wire.

A $\mathrm{10\ m}$ long conductor is formed into a circle in the $\mathrm{xy}$-plane. A uniform magnetic field  $\vec{B}=4\hat{i}+19\hat{j}+24\hat{k}\ \mathrm{T}$exists in the area of the conductor. Find the magnetic flux through the looped conductor.

The magnetic field strength within a long solenoid is $\mathrm{B=4.0\ t\ T}$, where $t$ is time in seconds. If the radius of the solenoid is $\mathrm{1.0\ cm}$, what is the magnitude of the non-Coulomb force that acts on a $\mathrm{6.0\ C}$ charge a distance $\mathrm{6.0\ cm}$ from the axis of the solenoid?

A uniform electric field is oriented so that it makes a $\mathrm{60.0{}^\circ }$ angle with the normal to a $\mathrm{1.0\ }{\mathrm{m}}^{\mathrm{2}}$ surface. The electric field is changing with respect to time by a rate of $\mathrm{6\ V/m . \mu s}$. Find the induced displacement current through the surface.

An electron and a proton are both initially moving with the same speed and in the same direction at $90{}^\circ$ to the same uniform magnetic field. They experience magnetic forces, which are initially:
(a) identical 
(b) equal in magnitude but opposite in direction 
(c) in the same direction and differing in magnitude by a factor of $1840$ 
(d) in opposite directions and differing in magnitude by a factor of $1840$ 
(e) equal in magnitude but perpendicular to each other.
 

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.
 

An electron is travelling in the positive $x$ direction. A uniform electric field $\vec{E}$ is in the negative $y$ direction. If a uniform magnetic field with the appropriate magnitude and direction also exists in the region, the total force on the electron will be zero. The appropriate direction for the magnetic field is:
(a) the positive $y$ direction 
(b) the negative $y$ direction 
(c) into the page 
(d) out of the page 
(e) the negative $x$ direction
 

The diagrams show five possible orientations of a magnetic dipole $\vec{\mu }$ in a uniform magnetic field $\vec{B}$ . For which of these does the magnetic torque on the dipole have the greatest magnitude?
(a) I
(b) II
(c) III
(d) IV
(e) V

Two parallel wires, $\mathrm{4\ cm}$ apart, carry currents of $\mathrm{2\ A}$ and $\mathrm{4\ A}$ respectively, in opposite directions. The force per unit length of one wire on the other is:
(a) $1\times {10}^{-3}\mathrm{N/m}$, repulsive
(b) $1\times {10}^{-3}\mathrm{N/m}$, attractive
(c) $4\times {10}^{-5}\mathrm{N/m}$, repulsive
(d) $4\times {10}^{-5}\mathrm{N/m}$, attractive
(e) None of these.

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.
 

Magnetic field lines inside the solenoid shown are:
(a) clockwise circles as one looks down the axis from the top of the page 
(b) counterclockwise circles as one looks down the axis from the top of the page 
(c) toward the top of the page 
(d) toward the bottom of the page 
(e) in no direction since $\mathrm{B\ =\ }0$
 

A square loop of wire lies in the plane of the page. A decreasing magnetic field is directed into the page. The induced current in the loop
(a) counterclockwise 
(b) clockwise
(b) zero 
(c) up the left edge and from right to left along the top edge 
(d) through the middle of the page

A cylindrical region of radius $\mathrm{R\ =\ 3.0\ cm}$ contains a uniform magnetic field parallel to its axis. If the electric field induced at a point $\mathrm{R/2}$ from the cylinder axis is $\mathrm{4.5\times }{\mathrm{10}}^{\mathrm{-}\mathrm{3}}\mathrm{\ V/m}$ the magnitude of the magnetic field must be changing at the rate of:
(a) $0\ \mathrm{T/s}$
(b) $0.30\ \mathrm{T/s}$
(c) $0.60\ \mathrm{T/s}$
(d) $1.2\ \mathrm{T/s}$
(e) $2.4\ \mathrm{T/s}$
 

Faraday's law says that a changing magnetic field with time induces an emf and a current in the loop that is $\mathcal{E}=-d{\mathrm{\Phi }}_B/dt$, where the minus sign is indication of Lenz's law. This induced current implies the presence of an induced electric field $\vec{E}$  that is tangent to loop. Recall that the line integral of $\vec{E}.d\vec{s}$ over a specific path defines the potential electric. In this cases there is a closed loop and an induced electric field so the line integral of above quantity gives the induced emf in the loop i.e. $\mathcal{E}=\oint{\vec{E}.d\vec{s}}$. By equating these relations, we obtain an alternative and useful form of Faraday's law as 
\[\oint{\vec{E}.d\vec{s}}=-\frac{d{\mathrm{\Phi }}_B}{dt}\] 
Therefore, it is sufficient to evaluate both sides of above relation at the location of $R/2$. 
\[-\frac{d{\mathrm{\Phi }}_B}{dt}=\oint{\vec{E}.d\vec{s}}\] 
\[-\frac{d}{dt}\left(BA\right)=E\oint{ds}\] 
\[-\frac{dB}{dt}(\pi r^2)=E\left(2\pi r\right)\] 
\[\Rightarrow \left|\frac{dB}{dt}\right|=\frac{2E}{r}=\frac{2E}{\frac{R}{2}}=4\frac{\left(4.5\times {10}^{-3}\right)}{3\times {10}^{-2}}=0.6\, \frac{\mathrm{T}}{\mathrm{s}}\] 
In the second equality we have used the definition 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.
 

A sinusoidal electromagnetic wave with an electric field amplitude of $\mathrm{100\ V/m}$ is incident normally on a surface with an area of $1\ {\mathrm{cm}}^{\mathrm{2}}$ and is completely absorbed. The energy absorbed in $\mathrm{10\ s}$ is:
(a) $1.3\ \mathrm{mJ}$
(b) $13\ \mathrm{mJ}$
(c) $27\ \mathrm{mJ}$
(d) $130\ \mathrm{mJ}$
(e) $270\ \mathrm{mJ}$


Magnetic fields
Category : Magnetic fields

MOST USEFUL FORMULA IN MAGNETIC FIELDS:

Magnetic Force on a moving charge
\[\vec{F}=q(\vec v\times \vec B)\]
Magnetic force on a current-carrying wire $I$
\[\vec F=i\vec \ell \times \vec B\]
magnetic flux through a surface $S$
\[\Phi_B=\int{\vec B \cdot \hat n d\vec \ell}\]
Ampere's Law:
\[\oint{\vec B \cdot d\vec{\ell}}=\mu_0I_{enc}\]
Magnetic field of a straight current carrying wire
\[B=\frac{\mu_0}{2\pi}\frac{I}{r}\]
Poynting vector 
\[\frac {1}{\mu_0} \vec E \times \vec B\]


Number Of Questions : 40