Question

The magnetic field in a region is given by $B=2r$, where $r$ is the perpendicular distance from the $\text{x-axis}$. The direction of the magnetic field is along the positive $\text{z -axis}$. A conducting rod of length $1\text{m}$ lies parallel to the $\text{y-axis}$ as shown in the figure. If the rod moves with the velocity $v=8\text{m/s}$ along the positive x-axis, find the emf induced between the ends of the rod.

• A. $48\text{V}$
• B. $40\text{V}$
• C. $35\text{V}$
• D. $24\text{V}$

Answer 1

The correct option is B $40\text{V}$

Given:
Consider a small element of length $dr$ lying at a distance $r$ from the x-axis.


Emf developed across this element is,

$dE=Bvdr=2r\times \left(8\right)\times dr$

$\therefore$ Emf developed across the whole length of the rod is,

$E=\int dE$

$={\int }_2^{3}16rdr=16{\left[\frac{r^2}{2}\right]}_2^3$

$=8\times [9-4]=40\text{V}$

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