Question

A conducting rod $AC$ of length $4l$ 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. $V_A-V_C=4B\omega l^2$
• D. $V_C-V_O=\frac{9}{2}B\omega l^2$

Answer 1

Answer: (A)

Given, length of rod $=4l$; magnetic field $=B$
If $\omega =$ angular velocity $\omega =\frac{v}{l}(v=velocity).$
Now, small emf (d ) due to small element de is:
$dϵ=Bvdl=Bwldl$
On integrating it upto length ' $l$ ' of rod, the emf $(epsilon)$ will be: $\epsilon =B\omega {\int }_0^{l}ldl$
$\epsilon =\frac{B\omega l^2}{2}=$ potential difference b/w $A\&0$.
Thus, $V_A-V_0=\frac{B\omega l^2}{2}$