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A conducting rod of length $L$ lies in $X Y$-plane and makes an angle $30^{\circ}$ with $X$-axis. One end of the rod lies at origin initially. A magnetic field also exists in the region pointing along positive $Z$-direction. The magnitude of the magnetic field varies with $y$ as $B_0\left(\frac{y}{L}\right)^3$, where, $B_0$ is a constant. At some instant the rod starts moving with a velocity $v_0$ along $X$-axis. The emf induced in the rod is
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Verified Answer
The correct answer is:
$\frac{B_0 V_0 L}{64}$
Length along $Y$-axis would be $l \sin 30^{\circ}=\frac{l}{2}$
length $=L$
Velocity of rod $=v_0$, along $X$-axis
variable magnet is field, $B=B_0\left(\frac{y}{L}\right)^3$, along $Z$-axis.
Now, induced emf (formula)
$$
E=B l v
$$
where, $B, l_{\perp}$ and $v$ are perpendicular to each other, $l=L / 2$
Now, consider a small element of length $d y$ at a distance $y$ from origin along $Y$-axis, then emf of element
$$
d E=B(d y) v_0=B_0 \frac{y^3}{L^3} v_0 \cdot d y
$$
Emf of the rod,
$$
\begin{aligned}
& E=\frac{B_0 v_0}{L^3} \int_0^{L / 2} y^3 \cdot d y \\
& E=\frac{B_0 v_0}{L^3}\left\lceil\frac{y^4}{4}\right\rceil^{L / 2} \\
& E=\frac{B_0 v_0}{L^3}\left\lfloor\frac{(L / 2)^4}{4}\right\rfloor \Rightarrow E=\frac{B_0 v_0 L}{64}
\end{aligned}
$$
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