Question

A semi circular wire of radius R is rotated with angular velocity $\omega$ about an axis passing through one end and perpendicular to the plane of wire in a uniform magnetic field of magnitude $B_o$. $q$ is midpoint of the semicircle, then the ratio of potential difference between p and r to q and r is ($i.e{V}_{pr}:V_{qr})$

• A. $\sqrt{2}:1$
• B. 1:2
• C. $\sqrt{3}:\sqrt{2}$
• D. 2 :1

Answer 1

The correct option is D 2 :1

$ϵ=\frac{B_{0}w{l}_{eff}^2}{2}\Rightarrow V_{pr}=\frac{B_{0}w(2R{)}^2}{2}=2B_{0}w{R}^2{V}_{pq}=\frac{B_{0}w(\sqrt{2}R{)}^2}{2}=B_{0}w{R}^2$
So
$V_{qr}=V_{pr}-V_{pq}=2B_{0}w{R}^2-B_{0}w{R}^2=B_{0}w{R}^2\text{so}{V}_{pr}:V_{qr}=2:1$