Question

A copper rod $ab$ of length $l$, pivoted at one end $a$, rotates at constant angular velocity $\omega$, at right angle to a uniform magnetic field $B$. $c$ is the midpoint of the rod. What will be the ratio of induced emf across $ac$ to $cb$?

• A. $3:2$
• B. $2:3$
• C. $3:1$
• D. $1:3$

Answer 1

The correct option is D $1:3$


Potential difference across $a$ and $c$,

$V_{ac}=\frac{B\omega L^2}{2}=\frac{B\omega l^2}{8}[\because L=l/2]$

Similarly,

$V_{ab}=\frac{B\omega l^2}{2}$

Assuming, $V_b=0$,

$V_a=\frac{B\omega l^2}{2}$ and $V_c=\frac{3B\omega l^2}{8}$

So, $V_{cb}=V_c-V_b=\frac{3B\omega l^2}{8}-0=\frac{3B\omega l^2}{8}$

Therefore, the ratio of induced emf across $ac$ to $cb$,

$=\frac{B\omega l^2/8}{3B\omega l^2/8}=\frac{1}{3}=1:3$

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


$\text{Alternative Solution}:$

Let us take a small element $dx$ at the distance of $x$ from end $a$.



Then, its linear speed will be $\omega x$ and small emf induced in it,

$dE=B\left(\omega x\right)dx$

So,

$V_a-V_c={\int }_0^{l/2}B\left(\omega x\right)dx=\frac{B\omega }{2}[x^2{]}_0^{l/2}=\frac{B\omega l^2}{8}$

Also,

$V_c-V_b={\int }_{l/2}^{l}B\left(\omega x\right)dx=\frac{B\omega }{2}[x^2{]}_{l/2}^l=\frac{3B\omega l^2}{8}$

Therefore, the ratio of induced emf across $ac$ to $cb$,

$=\frac{B\omega l^2/8}{3B\omega l^2/8}=\frac{1}{3}=1:3$