Question

A metallic V shaped rod OAC is rotated with respect to one of its end in a uniform magnetic field, such that axis of rotation is parallel to the direction of magnetic field. Length of each arm of the rod is L and the angle between the arms is $60^{\circ }$ . P is the mid-point of section AC. The magnitude of the magnetic field is B. Then choose the incorrect relation.

• A. $V_A-V_0=\frac{\omega B{L}^2}{2}$
• B. $V_A-V_C=\frac{\omega B{L}^2}{2}$
• C. $V_C-V_P=\frac{\omega B{L}^2}{8}$
• D. $V_A-V_P=\frac{\omega B{L}^2}{8}$

Answer 1

Answer: (B)

$OA=OC=L;OP=\frac{\sqrt{3}L}{2}$
$V_A-V_0=\frac{\omega B{L}^2}{2}{V}_A-V_C=0$
$V_C-V_0=\frac{\omega B{L}^2}{2}{V}_P-V_0=\frac{\omega B}{2}\left(\frac{3L^2}{4}\right)=\frac{3\omega B{L}^2}{8}$
$V_C-V_P=\left(V_C-V_0\right)-\left(V_P-V_0\right)=\frac{\omega B{L}^2}{2}-\frac{3\omega B{L}^2}{8}=\frac{\omega B{L}^2}{8}$
$V_A-V_P=\left(V_A-V_0\right)-\left(V_P-V_0\right)=\frac{\omega B{L}^2}{2}-\frac{2\omega B{L}^2}{8}=\frac{\omega B{L}^2}{8}$