Question 36: Electromagnetic Induction

Question

A conducting square loop of side $L$, mass $M$ and resistance $R$ is moving in the $X Y$ plane with its edges parallel to the $X$ and $Y$ axes. The region $y \geq 0$ has a uniform magnetic field, $\vec{B}=B_0 \widehat{k}$. The magnetic field is zero everywhere else. At time $t=0$, the loop starts to enter the magnetic field with an initial velocity $v_0 \hat{\jmath} \mathrm{~m} / \mathrm{s}$, as shown in the figure. Considering the quantity $K=\frac{B_0^2 L^2}{R M}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:

Accepted Answer

Here, as the loop enters the field, the magnetic flux through the loop is changing. Changing magnetic flux, according to Faraday's law, induces an EMF in the loop. As the loop has some resistance
So the loop gets an induced current and
The magnetic force arises from the induced current interacting with the field.
Flux : $\phi=B_0$. (Area in Field)
$$ \Rightarrow \phi=B_0 \cdot L y $$
EMF: $\quad V=\frac{-d \phi}{d t}=-\frac{d}{d t}\left(B_0 L yight)$.
$$ \begin{aligned} & \Rightarrow \quad v=-B_0 L \frac{d y}{d t} \ & \Rightarrow \quad v=-B_0 L v \end{aligned} $$
current $(v>0$, so current direction based on Lenz's (law): $I=\frac{V}{R}=\frac{-B_0 v L}{R}$
Force: current in the bottom edge (in field) interacts with $\vec{B}$
so force on a segment: $F=B_0 I L$
$$ \begin{aligned} & \Rightarrow F=B_0\left(\frac{-B_0 v L}{R}ight) L \ & \Rightarrow F=\frac{-B_0^2 L^2 v}{R} \end{aligned} $$
$$ 	ext { Given, } K=\frac{B_0^2 L^2}{R M} $$
$$ \begin{aligned} 	ext { or } \varepsilon & =-B_0 v L \ I & =\frac{-B_0 v L}{R} \ F & =B_0 I L=B_0\left(\frac{-B_0 v L}{R}ight) L \end{aligned} $$
$$ \Rightarrow \frac{m d v}{d t}=\frac{-\left (B_0^2 L^2ight) v}{R} \quad\left[\begin{array}{ll} 	ext { as } & F=m a \ & =m \frac{d v}{d t} \end{array}ight] $$
$$ \begin{aligned} & \Rightarrow  	ext { } \frac{mvd v}{d x}=\frac{-\left(B_0^2 L^2ight)}{R} {v} \ & \Rightarrow \frac{d v}{d x}=-\frac{B_0^2 L^2}{M R} \ & \Rightarrow \int_{v_0}^v d v=-K \int_0^{x} dx\ & \Rightarrow v-v_0=-K x \ & 	ext { If } v_0=1.5 \mathrm{KL} 	ext { then for } x=L \end{aligned} $$
$$ \begin{aligned} & \Rightarrow v-1.5 \mathrm{KL}=-K L \ & \Rightarrow \quad v=1.5 \mathrm{KL}-\mathrm{KL}=0.5 \mathrm{KL} \ & \Rightarrow \quad v=0.5 \mathrm{KL} \end{aligned} $$
So, the loop will have some velocity, and it will not stop.
If the loop completely gets inside then $\Phi$ becomes constant so $\frac{d \Phi}{d t}=\varepsilon=0, \quad I=0$
i.e., no current flows and no force acts.
if $v_0=\frac{K L}{10}$ then for $v=0$
$$ \begin{aligned} & 0+\frac{K L}{10}=-k x \Rightarrow x=\frac{L}{10} \ & \frac{d v}{d t}=\frac{-B_0^2 L^2}{M R} v \ & \Rightarrow \int_{v_0}^v \frac{d v}{v}=-k \int_0^t d t \ & \Rightarrow[\ln |v|]_{v_0}^v=-k t \ & \Rightarrow \ln \left|\frac{v}{v_0}ight|=-k t \ & \Rightarrow t=\frac{1}{K} \ln \left|\frac{v_0}{v}ight| \ & 	ext { for } v=0, t=\frac{1}{k} \ln (\infty)=\infty \end{aligned} $$
So it will never stop.
If $v_0=3 \mathrm{KL}$ then
$v$ at $x=L$,
$v-3 K L=-K L$
$$ \Rightarrow v=3 K L-K L=2 K L $$
So time, $t=\frac{1}{k} \ln \left|\frac{v_0}{v}ight|$
$$ \begin{aligned} & \quad t=\frac{1}{k} \ln \left|\frac{3 K L}{2 K L}ight| \ & \Rightarrow t=\frac{1}{k} \ln \left(\frac{3}{2}ight) \end{aligned} $$
Hence, Options (B) and (D) are correct.

Answer

If $v_0 = 1.5KL$, the loop will stop before it enters completely inside the region of magnetic field.

Answer

When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.

Answer

If $v_0 = \frac{KL}{10}$, the loop comes to rest at $t = \left(\frac{1}{K}\right) \ln\left(\frac{5}{2}\right)$.

Answer

If $v_0 = 3KL$, the complete loop enters inside the region of magnetic field at time $t = \left(\frac{1}{K}\right) \ln\left(\frac{3}{2}\right)$.

Electromagnetic Induction

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