Question

A conducting rod $\text{AC}$ of length $4l$ is rotated about a point $\text{O}$ in a uniform magnetic field $B$ as shown in the figure. If $AO=l$ and $OC=3l$ then

• A. $V_A-V_O=\frac{B\omega l^2}{2}$
• B. $V_O-V_C=\frac{9B\omega l^2}{2}$
• C. $V_A-V_C=0$
• D. $V_A-V_C=4B\omega l^2$

Answer 1

Answer: (B), (D)

$V_A-V_C=4B\omega l^2$

Emf developed across the part $\text{AO}$,

$E_{OA}=\frac{B\omega l^2}{2}$

According to the right-hand rule, point $\text{A}$ will be at lower potential and point $\text{O}$ will be at higher potential.

$\therefore V_0-V_A=\frac{B\omega l^2}{2}$

Similarly,

$E_{OC}=V_0-V_C=\frac{B\omega (3l{)}^2}{2}$

$V_0-V_C=\frac{9B\omega l^2}{2}$


$\therefore$ Potential difference between points $\text{C}$ and $\text{A}$ is:

$V_A-V_C=E_{OC}-E_{OA}$

$=\frac{9B\omega l^2}{2}-\frac{B\omega l^2}{2}=4B\omega l^2$

Therefore, options $\left(\text{B}\right)$ and $\left(\text{D}\right)$ are the correct answers.

$\text{Alternate Solution}:$ Consider an element at distance $x$ from the centre of the rod, as shown in the figure. Emf induced across the element due to rotation, $E=Bvl=B\left(\omega x\right)dx[\because v=\omega x]$ Now, the total emf developed across the part $\left(\text{AO}\right)$ or potential difference between the ends $\text{A}$ and $\text{O}$, $E_1=V_A-V_O={\int }_{-l}^{0}B\omega xdx$ $=B\omega {\left[\frac{x^2}{2}\right]}_{-l}^0=\frac{B\omega }{2}[0^2-(-l{)}^2]=-\frac{B\omega l^2}{2}$ $\therefore V_O-V_A=\frac{B\omega l^2}{2}$ Similarly, for the part $\text{O}C$, $E_1=V_O-V_C={\int }_0^{3l}B\omega xdx=\frac{9B\omega l^2}{2}$ Potential difference between $textA$ and $\text{C}$, $V_A-V_C={\int }_{-l}^{3l}B\omega xdx=4B\omega l^2$ Therefore, options $\left(\text{B}\right)$ and $\left(\text{D}\right)$ are the correct answers.