Question

A conducting rod $AC$ of length $4 l$ is rotated about appoint $O$ in a uniform magnetic field $\vec{B}$ directed into the paper. $AO=l$ and $OC=3l$ then:

• A. $V_{A}-V_{O}=\frac{B \omega l^{2}}{2}$
• B. $V_{O}-V_{C}=\frac{9}{2} B \omega l^{2}$
• C. $V_{A}-V_{C}=4 B \omega l^{2}$
• D. $V_{C}-V_{O}=\frac{9}{2} B \omega l^{2}$

Answer 1

Answer: (A)

$V_{A}-V_{O}=\frac{B \omega l^{2}}{2}$

Answer 2

Given, length of rod $=4 l$; magnetic field $=B$ If $\omega=$ angular velocity $\omega=\frac{v}{l} \quad(v=v e l o c i t y) .$ Now, small emf (d ) due to small element de is: $d \epsilon=B vdl =B wl d l$ On integrating it upto length ' $l$ ' of rod, the emf $(epsilon)$ will be: $\varepsilon=B \omega \int_{0}^{l} l d l$ $\varepsilon=\frac{B \omega l^{2}}{2}=$ potential difference b/w $A \& 0$. Thus, $V_{A}-V_{0}=\frac{B \omega l^{2}}{2}$