Question

The given figure shows a conductor $\text{OA}$ of length $l$ placed along $\text{y}$-axis with one end at the origin. In this region, a non-uniform magnetic field exists having magnitude $B=B_0(1+\frac{y^2}{l^2})$ along $\text{+z}$- direction. If the conductor $\text{OA}$ starts translating with velocity $\overrightarrow{v}=v_0\hat{i}$, then find the emf induced in the conductor.

• A. $\frac{B_0{v}_{0}l}{2}$
• B. $B_0{v}_{0}l$
• C. $\frac{4}{3}{B}_0{v}_{0}l$
• D. Zero

Answer 1

The correct option is C $\frac{4}{3}{B}_0{v}_{0}l$

Let us consider a small element of length $dy$ at a distance $y$ from the origin.


Motional emf induced in this element is,

$dE=B{v}_{0}dy=B{v}_0(1+\frac{y^2}{l^2})dy$

$\therefore$ Total emf induced across $OA$ is,

$E_{OA}=\int dE={\int }_0^l{B}_0{v}_0(1+\frac{y^2}{l^2})dy$

$=B_0{v}_0{[y+\frac{y^3}{3l^2}]}_0^l$

$\therefore E_{OA}=\frac{4}{3}{B}_0{v}_{0}l$

Hence, $\left(C\right)$ is the correct answer.