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

• Correct answer
  A

  $\frac{B_0 V_0 L}{64}$
• B

  $\frac{B_0 V_0 L}{16}$
• C

  $B_0 V_0 L$
• D

  $64 B_0 v_0^L$

Solution

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, $$ \begin{aligned} & E=\frac{B_0 v_0}{L^3} \int_0^{L / 2} y^3 \cdot d y \\ & E=\frac{B_0 v_0}{L^3}\left\lceil\frac{y^4}{4}\right\rceil^{L / 2} \\ & E=\frac{B_0 v_0}{L^3}\left\lfloor\frac{(L / 2)^4}{4}\right\rfloor \Rightarrow E=\frac{B_0 v_0 L}{64} \end{aligned} $$

Hence, the correct answer is option: 'A'.