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A magnetic field $\mathrm{B}=\mathrm{B}_0 \sin ((\omega \mathrm{t}) \hat{\mathrm{k}}$ covers a large region where a wire $\mathrm{AB}$ slides smoothly over two parallel conductors separated by a distance $\mathrm{d}$ (figure). The wires are in the $\mathrm{x}-\mathrm{y}$ plane.
The wire $A B$ (of length $d$ ) has resistance $R$ and the parallel wires have negligible resistance. If $\mathrm{AB}$ is moving with velocity $\mathrm{v}$, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?
The wire $A B$ (of length $d$ ) has resistance $R$ and the parallel wires have negligible resistance. If $\mathrm{AB}$ is moving with velocity $\mathrm{v}$, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?
Solution:
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Verified Answer
Consider the parallel wires at $y=0$. i.e., along $\mathrm{x}$-axis and $\mathrm{y}=\mathrm{d}$.
At $\mathrm{t}=0, \mathrm{AB}$ has $\mathrm{x}=0$, i.e., along $\mathrm{y}$-axis and moves with a velocity $v$. Let at timet, wire is at $\mathrm{x}(\mathrm{t})=\mathrm{vt} \hat{\mathrm{i}}$, So, the motional emf across $\mathrm{AB}$ is
$$
\varepsilon_1=\left(B_0 \sin \omega t\right) v d(-\hat{j})=-\left(B_0 \sin \omega t\right) v d
$$
emf due to change in field (along $\mathrm{OBAC}$ )
$$
\varepsilon_2=-\mathrm{B}_0 \omega \cos \omega \mathrm{t} x(\mathrm{t}) \mathrm{d}
$$
Total emf in the circuit $=$ emf due to change in field (along $\mathrm{OBAC})$ + the motional emf across $\mathrm{AB}=-\left[\mathrm{B}_0 \omega \cos \omega \mathrm{t}(\mathrm{x}(\mathrm{t}) \mathrm{d}\right.$ $\left.+\left(\mathrm{B}_0 \sin \omega \mathrm{t}\right) \mathrm{vd}\right]$ $\varepsilon=-\mathrm{B}_0 \mathrm{~d}[\omega \mathrm{x} \cos (\omega \mathrm{t})+\mathrm{vsin}(\omega \mathrm{t})]$
Electric current in clockwise direction is
$$
\mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B}_0 \mathrm{~d}}{\mathrm{R}}(\omega x \cos \omega \mathrm{t}+\mathrm{v} \sin \omega \mathrm{t})
$$
The force acting on the conductor is
$$
\mathrm{F}=\mathrm{i} / \mathrm{B} \sin 90^{\circ}=\mathrm{i} / \mathrm{B}
$$
Substituting the values, we get
Then force needed along $\mathrm{i}$
$\begin{aligned}=& \frac{B_0 d}{R}(\omega x \cos \omega t+v \sin \omega t) \times d \times B_0 \sin \omega t \\ F &=\frac{B_0^2 d^2}{R}(\omega x \cos \omega t+v \sin \omega t) \sin \omega t \end{aligned}$
At $\mathrm{t}=0, \mathrm{AB}$ has $\mathrm{x}=0$, i.e., along $\mathrm{y}$-axis and moves with a velocity $v$. Let at timet, wire is at $\mathrm{x}(\mathrm{t})=\mathrm{vt} \hat{\mathrm{i}}$, So, the motional emf across $\mathrm{AB}$ is
$$
\varepsilon_1=\left(B_0 \sin \omega t\right) v d(-\hat{j})=-\left(B_0 \sin \omega t\right) v d
$$
emf due to change in field (along $\mathrm{OBAC}$ )
$$
\varepsilon_2=-\mathrm{B}_0 \omega \cos \omega \mathrm{t} x(\mathrm{t}) \mathrm{d}
$$
Total emf in the circuit $=$ emf due to change in field (along $\mathrm{OBAC})$ + the motional emf across $\mathrm{AB}=-\left[\mathrm{B}_0 \omega \cos \omega \mathrm{t}(\mathrm{x}(\mathrm{t}) \mathrm{d}\right.$ $\left.+\left(\mathrm{B}_0 \sin \omega \mathrm{t}\right) \mathrm{vd}\right]$ $\varepsilon=-\mathrm{B}_0 \mathrm{~d}[\omega \mathrm{x} \cos (\omega \mathrm{t})+\mathrm{vsin}(\omega \mathrm{t})]$
Electric current in clockwise direction is
$$
\mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B}_0 \mathrm{~d}}{\mathrm{R}}(\omega x \cos \omega \mathrm{t}+\mathrm{v} \sin \omega \mathrm{t})
$$
The force acting on the conductor is
$$
\mathrm{F}=\mathrm{i} / \mathrm{B} \sin 90^{\circ}=\mathrm{i} / \mathrm{B}
$$
Substituting the values, we get
Then force needed along $\mathrm{i}$
$\begin{aligned}=& \frac{B_0 d}{R}(\omega x \cos \omega t+v \sin \omega t) \times d \times B_0 \sin \omega t \\ F &=\frac{B_0^2 d^2}{R}(\omega x \cos \omega t+v \sin \omega t) \sin \omega t \end{aligned}$
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