Question

A frame ABCD is rotating with angular velocity $\omega$ about an axis which passes through the point $O$ perpendicular to the plane of the paper as shown in figure. A uniform magnetic field $\overrightarrow{B}$ is applied into the plane of the paper in the region as shown. Match the following.

$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A) Potential difference between A and O}& \text{(p) zero}\\ \text{(B) Potential difference between O and D}& \text{(q)}B\omega L^2/2\\ \text{(C) Potential difference between C and D}& \text{(r)}B\omega L^2\\ \text{(D) Potential difference between A and D}& \text{(s) constant}\end{array}$

• A. A – (p,s), B – (p,s), C – (q,s), D – ( r,s)
• B. A – (r,s), B – (p,s), C – (q,s), D – ( r,s)
• C. A – (r,s), B – (r,s), C – (q,s), D – ( p,s)
• D. A – (q,s), B – (p,s), C – (r,s), D – ( r,s)

Answer 1

The correct option is C A – (r,s), B – (r,s), C – (q,s), D – ( p,s)

We know that motional EMF in a rotating body is given by
$V=\frac{1}{2}B\omega L^2$

Replacing the rods with equivalent cells as shown, the potential difference between various points is:

For AO and DO
$L=L\sqrt{2}$
$V_{AO}=V_{OD}=\frac{1}{2}B(L\sqrt{2}{)}^2\omega =B\omega L^2$
(A and B)

From this we get
$\Rightarrow V_{AD}=0$ (D)

$V_A-B{L}^2\omega =V_0$ ...(i)
$V_B-\frac{1}{2}B{L}^2\omega =V_0$ ....(ii)
Solving (i) and (ii) we get
$\Rightarrow V_A-\frac{1}{2}B{L}^2\omega =V_B\Rightarrow V_A-V_B=\frac{1}{2}B{L}^2\omega =V_C-V_D$
(C)