Solved Problems

(a) The magnetic flux through any loop is defined as ${\Phi }_M=\vec{B}\cdot\vec{A}$, so \[\Phi_M=3t^2(-\hat{k})\cdot 2(-\hat{k})=6t^2\] 

(b) Faraday's law states that the induced emf in any closed loop (circuit) is the negative of the time rate of change of the magnetic flux through it i.e. \[\mathcal E=-\frac{d \Phi_M}{d t}=-12t\] 
So the magnitude of induce current in the loop is $I_{ind}=|\mathcal E|/R$ so \[I_{ind}=\frac{12t}{3}=4{\left.t\right|}_{t=2{\rm s}}=8\ {\rm A}\] Since the magnetic field $B$ is increasing, the magnetic flux through the loop is also increasing. 

Lenz's law tells us that the direction of the induced current is such that the induced magnetic field opposes the original flux change. 

Thus, the induced current must be counter-clockwise.

(c) see part (b)
 

(a) From the Ohm's law the current through a metal wire is $I=({\rm potential\ drop)/R}$ so first using the Faraday's law find the emf induced in this loop and then substitute it in the Ohm's law. By definition, the emf is  
\[{\mathcal E}=-\frac{\Delta \Phi_M}{\Delta t}\ ,\ {\rm where}\ \Phi_M=BA\] 
Since there is an increasing magnetic field, the change of the flux is 
\[\Phi_8=B_8A=B_8\pi r^2=(1)\pi{\left(0.40\right)}^2=0.5024\ {\rm out\ \ \ or\ -0.5024\ }\left({\rm T.}{{\rm m}}^{{\rm 2}}\right){\rm \ \ in}\] 
\[\Phi_0=B_0A=3\pi{\left(0.40\right)}^2=1.5072\ {\rm into}\] 
\[\Delta \Phi_M=\Phi_8-\Phi_0=\left(-0.5024\ {\rm in}\right){\rm -}\left(1.5072\ {\rm in}\right){\rm =-2.01\ }\left({\rm T.}{{\rm m}}^{{\rm 2}}\right)\ \ {\rm into}\] 
In above, we have assigned a convention that into (out of) the page to be negative(positive). 

The flux is decreasing, so by Lenz's law, the induced current must be in a direction to produces an induced field to oppose with this decreasing that is it must be clockwise!
\[{\mathcal E}=-\frac{\Delta \Phi_M}{\Delta t}=-\frac{-2.01}{8-0}=0.2512\ {\rm V}\] 
\[{\rm Oh}{{\rm m}}^{{\rm '}}{\rm s\ law}\ :\ I_{induced}=\frac{{{\mathcal E}}_{av}}{R}=\frac{0.2512}{15}=0.01675\ {\rm A}\] 
(b) Stated above. (clock wise) 

(c) Since we don’t know the magnitude of the flux through the loop for all time, so we must use the word average.

(a) ${\Phi }_M=\vec{B}.\vec{A}=BA\,{\cos  37{}^\circ \ }=10\times \pi{\left(0.18\right)}^2\times \left(0.8\right)=0.813\ {\rm Wb}$

(b) 
\[\frac{\Delta {\Phi }_M}{\Delta t}=\frac{{\Phi }_M\left(t=0.1\right)-{\Phi }_M\left(t=0\right)}{0.1-0}=\frac{100\times 0.813-0}{0.1}=813\ {\rm Wb/s}\] 
Note: the total flux through the entire coil is ${\Phi }_{tot}=N\Phi $
(c) ${\mathcal E}=-\frac{\Delta {\Phi }_M}{\Delta t}\to \ \left|{\mathcal E}\right|=813{\rm V}$

(d) $I_{ind}=\frac{{\mathcal E}}{R}=\frac{813}{375}=2.16{\rm A}$

(e) By Lenz's law, due to the increasing flux through the coil, the induced magnetic field produced by this current should be in direction to oppose the external magnetic field so using right hand rule the induced current is clockwise. 

(f) Because the area of the coil is constant and during the time $\left[0,0.15{\rm s}\right]$ the variation of the magnetic field is zero so $\Delta {\Phi }_M=0\Rightarrow {\mathcal E}=0$   
 

(a) Due to the change of flux (as the slide moves the area of the loop varies), there is an induced current in the slide wire. By definition,
\begin{align*} \mathcal E&=-\frac{d}{dt}\Phi_B \\\\ &=-\frac{d}{dt}\left(\vec{B}\cdot \hat{n}dA\right) \\\\ &=-B\frac{dA}{dt} \quad \text{where} A=Lx \end{align*} In the above $\hat{n}$ is a unit vector perpendicular to the area of the surface, which in this case, makes an angle of $90^\circ$ with the surface of the loop. By definition of scalar product, we have $\vec{B}\cdot \hat{n}=B\cos\theta=B\cos 0=B$. Hence, the induced emf in the loop is found to be \[\mathcal E=-BL\frac{dx}{dt}=-BLv\] 
From Ohm's law formula, we can obtain the induced current in the slide wire as below: \[i_{ind}=\frac{\mathcal E}{R_{wire}}=\frac{BLv}{R}\] 
In the above, the minus sign has been omitted since this negative is a sign of Lenz's law! 
When the wire moves to the right, the flux through the loop is increasing, so by Lenz's law; the direction of induced current must be counterclockwise that is the current in the wire must be upward, say in the positive $y$ direction (or $\hat{j}$). 

On the other hand, we are in a situation where there is a current-carrying wire in a uniform magnetic field. In these cases, the magnetic force on the wire is determined by the formula $\vec{F}=i\vec{L}\times \vec{B}$ where $\vec{L}$ is a vector that its magnitude determines the straight length of the wire is in the magnetic field, and its direction shows the direction of the current in the wire. \begin{align*}\vec{F}&=i\vec{L}\times \vec{B} \\ &=i(L\hat{j}) \times B(-\hat{j}) \\ &= iLB\,(\hat{j}\times (-\hat{k})) \\ &=-iLB\hat{i}\end{align*} Substituting the induced current found in above into this expression, we will have \[\vec{F}=\frac{B^2L^2v}{R}(-\hat{i})\] $-\hat{i}$ indicates the direction of the magnetic force on this wire that is to the left. 
 
(b) Use Newton's second law and definition of the instantaneous acceleration and some manipulation to find instantaneous velocity of the wire as follows:
\begin{gather*}\vec{F}=m\vec{a}=m\frac{dv}{dt}\\ \\-\frac{B^2L^2v}{R}=m\frac{dv}{dt}\\ \\\to -\frac{B^2L^2}{Rm}dt=\frac{dv}{v}\end{gather*} By integrating both sides, we have \begin{gather*} \int^v_{v_0}{\frac{dv}{v}}=-\int{\frac{B^2L^2}{mR}dt}\\ \\ \to \quad {\ln  \frac{v}{v_0}\ }=-\frac{B^2L^2}{mR}t \\ \\ \Longrightarrow \quad v=v_0\,{\exp  \left(-\frac{B^2L^2}{mR}t\right)}\end{gather*}
We can check its correctness by substituting the initial and final time into it:
$t=0\to v=v_0\ $and $t\to \infty \ \Longrightarrow v=v_0\,e^{-\infty }\to 0$
Now use the definition of the instantaneous velocity to determine the desired distance $x$ as below \begin{align*} v&=\frac{dx}{dt} \\ \\ \to x &= \int{v\left(t\right)dt}\\ \\&=v_0\int{{\exp  \left(-\frac{B^2L^2}{mR}t\right)}dt}\\\\ &=-\frac{mRv_0}{B^2L^2}{\left.{\exp  \left(-\frac{B^2L^2}{mR}t\right)\ }\right|}^t_0\\ \\&=\frac{mRv_0}{B^{2}L^2}\left(1-e^{-\frac{B^{2}L^2}{mR}t}\right)\end{align*}
In the limit that $t$ becomes infinitesimally small $t\to \infty$ we get  $x(t)=\frac{mR}{B^{2}L^{2}}v_0$.

Lenz's law determines the direction of the induced current. This law states that the direction of the induced current must be such that the produced magnetic field by this current opposes the original flux change. 
(a) Since the area of the loop is constant and $\frac{dB}{dt}>0$, so by the definition of the magnetic flux i.e. ${\Phi }_M=\vec{B}.\hat{n}\ dA$, there is an increasing magnetic flux. By Lenz's law, the induced magnetic field must be opposite to the original field (downward), thus the induced current must be counterclockwise to produce an upward magnetic field. 
(b) With the similar reasoning as above, by Lenz's law the induced current must be in a direction to cancel out the decreasing downward flux so the current is clockwise.
(c) If $\frac{dB}{dt}=0$ then from $\frac{d}{dt}\Phi =\frac{d}{dt}\vec{B}.\hat{n}dA$, there is no change in flux and thus there is no induced current.   

(a) Use the definition of the instantaneous velocity that is $v=dz/dt$.  Since 
\begin{gather*}\vec{V}(t)=V(t)\hat{z} \\\\ \text{where} \quad V(t)<0 \quad \text{and} \quad z(t)>0 \\ \\  \text{then} \quad  \frac{d}{dt}z(t)=V(t)<0 \end{gather*}
(b) \begin{align*} \Phi_B&=\int{\vec{B}\cdot \hat{n}\, dA} \\\\ &=\int{BdA\,\cos 0} \\\\ &=B\int{da}=B(rectangular\ loop\ area) \\\\ &=Bwz(t)>0 \end{align*}
(c) Applying magnetic flux formula, we have \begin{align*}\frac{d\Phi_B}{dt}&=\frac{d}{dt}\left(Bwz\left(t\right)\right)\\\\&=wB\frac{d}{dt}z\left(t\right)\\\\&=BwV\left(t\right)<0\end{align*}
(d) Using Faraday's law of induction, we have  \[{\mathcal E}=-\frac{d\Phi_B}{dt}=-BwV\left(t\right)>0\] Therefore, using Ohm's law, we get \[I=\frac{\left|{\mathcal E}\right|}{R}=\frac{wB\left|V\left(t\right)\right|}{R}\]
Due to Lenz's law, the induced current must be counterclockwise so that the magnetic field produced by this current opposes the decreasing flux.
(e) As noted above, by using the right-hand rule $B'$ is out of the page.

(f) Since there is an induced current in the loop and lies in an external magnetic field so a magnetic force ${\vec{F}}_m=I_{ind}\vec{L}\times {\vec{B}}_{ext}$ exerts on it in the up direction. Thus, its magnitude is
\[L\bot B\Rightarrow \ \ F_m=I_{ind}wB_{ext}\] 
see figure below.
(g) When net force on the object is zero, the object has zero acceleration, in the other words, this object is moving at its terminal velocity. In this problem, two forces, gravity and magnetic, act on the upper side of the loop. So the magnetic force acts as the drag force! Write out this conditions, we obtain   
\[\Sigma F_y=ma_z=0\Rightarrow F_m\hat{j}+mg\left(-\hat{j}\right)=0\] 
$F_m=F_g\to \ \ I_{ind}wB_{ext}=mg\ $where the induced current is found in part(d) 
\[I_{induced}=\frac{wB_{ext}v_{terminal}}{R}\] 
\[\Longrightarrow w^2B^2\frac{v_{ter}}{R}=mg\ \ \Rightarrow \ \ v_{ter}=\frac{Rmg}{w^2B^2}\] 

Apply a current $I_1$ in the straight wire and find the flux passing through the loop.
From Ampere's law $\oint{\vec B.d\vec \ell}=\mu_0I\ $ the magnetic field produced by straight wire at distance $r$ is $B_1=\frac{\mu_0I_1}{2\pi r}$.

Using the definition of flux passing through the loop, we have 
\[{\Phi }_{21}=\int {\vec{B}}_1.\hat{n}\ da=\frac{\mu_0I_1}{2\pi}\int^R_0{\rho d\rho}\int^{2\pi}_0{\frac{d\theta}{r}}\] 
Where $r=d+\rho\,{\cos \theta\ }$ and $da=\rho d\rho \,d\theta$ is the area element in polar coordinates.(see figure). ${\Phi }_{21}$ is the flux produced by the current $I_1$ and passed the loop. 
Using the hint $\int^{2\pi}_0{\frac{d\theta}{d+\rho\,{\cos \theta\ }}=2\pi/\sqrt{d^2-\rho^2}}$ ,one can show that \[{\Phi }_{21}=\mu_0I_1\ \int^R_0{\frac{\rho d\rho}{\sqrt{d^2-\rho^2}}}\\] For determining the integral above we use the change of variables
let $u\equiv d^2-\rho^2$ , $\rho d\rho=-\frac{du}{2}$ \begin{align*}{\Phi }_{21}&=\mu_0I_1\ \int^R_0{\frac{\rho d\rho}{\sqrt{d^2-\rho^2}}}\\ \\&=-\frac{1}{2}\mu_0I_1 \int^{d^{2}-R^{2}}_{d^{2}}{\frac{du}{\sqrt{u}}}\\ \\&=\mu_0I_1 \left(d-\sqrt{d^2-R^2}\right)\end{align*} Thus, the mutual inductance, by definition, is 
\[M_{21}=\frac{{\Phi }_{21}}{I_1}=\mu_0\left(d-\sqrt{d^2-R^2}\right)\]

(a) The power supplied or dissipated in an electric circuit is $P=RI^2={\mathcal E}I=\frac{{{\mathcal E}}^2}{R}$. Therefore, we must first find the emf of the circuit and then substituting it into one of the above relations. Since we have a changing flux, so an emf produce in the loop. By definition its magnitude is 
\begin{align*}{\mathcal E}&=-\frac{d{\Phi }_B}{dt}\\&=-B\frac{dA}{dt}\\\&=-BL\frac{d}{dt}\left(vt\right)\\&=-BvL\\ \\ P_{diss}&=\frac{{{\mathcal E}}^2}{R}=\frac{B^2v^2L^2}{R}\end{align*}

(b) As soon as the current $I$ starts flowing around the loop, a magnetic force $F$ exerts on the wire that is found by $F=ILB=\frac{BvL}{R}.LB=B^2L^2v/R$. Now by definition, the mechanical power is \[P_{mech}=\frac{work}{time}=\frac{dW}{dt}=\frac{d}{dt}\left(\vec{F}.\vec{x}\right)=\vec{F}.\frac{d\vec{x}}{dt}=\vec{F}.\vec{v}\] Substituting the force exerted into this relation, we obtain
\[P_{mech}=Fv=\frac{B^2L^2v^2}{R}\] 
Note that the $P_{diss}=P_{mech}$ a consequence of conservation of energy!
 

(a) The direction of the induced current predicted by Lenz's law. According to this law, the direction of the induced current is such that the magnetic field produced by it, opposes the change in flux of the original magnetic field. Because the original magnetic field is decreasing so a clockwise induced current compensates for the original decreasing magnetic flux passing through the loop. 
(b) A change in magnetic flux passing through any closed loop produces an induced emf and current in the loop. This is Faraday's law of magnetic induction. In mathematical language is 
\[{\mathcal E}=-\frac{d{\Phi }_M}{dt}=-\frac{d}{dt}(\vec{B}.\vec{A})\] 
Where ${\Phi }_M$ is the magnetic flux. If the loop has $N$ turns then the total flux passing through them is $N$ times the flux through each of them.

First, use Faraday's law equation to find the induced emf created in the wire then use the Ohm's law and determine the magnitude of the induced current. \begin{gather*}{\mathcal E}=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}=-N\left(\pi r^2\right)\frac{dB}{dt}\\ \\{\mathcal E}=-100\left(\pi \times 3^2\right)(\underbrace{-1}_{decreasing})=+900\pi \ {\rm V}\end{gather*}
Then, using Ohm's law gives \[I_{ind}=\frac{{\mathcal E}}{R}=\frac{900\pi}{10}=90\pi\ {\rm A}\] 
 

We have two inductances, one is the mutual inductance which is present in two separate and independent circuits, and the other is self-inductance that occurs only in a single isolated circuit.

When a current is present in a circuit, it creates a magnetic field that causes a magnetic flux through that circuit. This flux changes when the current changes. Recall that Faraday's law states that the change in flux gives rise to an induced emf. Thus any circuit that carries a varying current has an induced emf that is produced by the variation in its own magnetic field.

The magnetic flux of $\vec{B}$ through the circuit is proportional to $I$ that is 
\[{\Phi }_m=LI\] 
Where $L$ is a proportionality constant which is called self-inductance of the circuit. The SI unit of it is henry $H$.

 In a varying current, the flux changes and so using Faraday's law we have
\[{\mathcal E}=-\frac{d{\Phi }_m}{dt}=-L\frac{dI}{dt}\] 

In this problem, $L=0.25\ {\rm H}$, $t=\frac{1}{16}\ {\rm s}$ so the induced emf is
\begin{align*}{\mathcal E}&=-L\frac{\Delta I}{\Delta t}\\ \\&=-L\frac{I_2-I_1}{\Delta t}\\\\&=-0.25\frac{0-2}{\frac{1}{16}}\\\\&=8\quad {\rm V}\end{align*}
 

This configuration is into a uniform magnetic field, but its surface is changing. So there is a changing magnetic flux.

By Faraday's law, a changing magnetic flux produces an induced emf and induced current in the circuit.

First, find this induced emf, then use Ohm's law to determine the induced current. Faraday's law equation is \[{\mathcal E}=-\frac{d{\Phi }_M}{dt}=-\frac{d}{dt}\int{\vec{B}\cdot\hat{n}dA}\] 
$B$ is uniform so we can factor it out of the integral. The remaining integral is the area of the changing shape which is easier to calculate it in the polar coordinate.

In this coordinate, the formula for finding this area is \[A=\int^\beta_\alpha{\frac{1}{2}r^2d\theta}\] Where $r$ is the distance to the center, which in this case, $r=D$ is the radius of the circle.

Therefore, the magnitude of the induced emf is \begin{align*}|{\mathcal E}|&=B\frac{d}{dt}\int{dA}\\ \\&=B\frac{d}{dt}\int^\theta_0{\frac{1}{2}D^2d\theta}\\\\&=\frac{B}{2}D^2\underbrace{\frac{d\theta}{dt}}_{\omega}\\ \\ \Rightarrow \quad  {\mathcal E}&=\frac{B}{2}D^2\omega\end{align*}
Using Ohm's law, the induced current will be \[I_{ind}=\frac{{\mathcal E}}{R}=\frac{1}{2R}BD^2\omega\] 
 

(a) Recall that there is a force acted on a current carrying wire or bar of length $\vec{l}$ in a magnetic field $\vec{B}$ as $\vec{F}=i\vec{l}\times \vec{B}$.

Here, the bar is perpendicular to the magnetic field, so the magnitude of the force is $F=ilB$. Using Newton's second law $F=ma$, and kinematic equation $v^2-v^2_0=2a\Delta x$, we can find the current through the bar
\[ilB=m\frac{\left(v^2-v^2_0\right)}{2\Delta x}\to i=\frac{m\left(v^2-v^2_0\right)}{2lB\Delta x}\] 
The bar is initially at rest, $v_0=0$ and its final velocity after travelling distance $1\ {\rm m}$  is $v=25\ {\rm m/s}$. by substituting these into above, we obtain the current 
\begin{align*}i&=\frac{\left(1.5\times {10}^{-3}\ {\rm kg}\right)\left({\left({\rm 25}\right)}^{{\rm 2}}-0^2\right)}{2\left(0.24\right)\left(1.8\right)\left(1\right)}\\ \\&=1.08\ {\rm A}\end{align*}

(b) Acceleration of any object is in direction of the force $\vec{a}=\vec{F}/m$, so for the force acting on the bar to be in the direction shown in the figure, using right hand rule, the magnetic field must be into the page.