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A semi circular rigid metallic wire, of radius R is rotating with constant angular speed $ω$ about an axis passing through point O and perpendicular to the plane of the semicircular wire in a region of uniform magnetic field ${'B}_{0}'$ as shown then

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Potential difference between points O and B is

\frac{{3B}_{0} R^{2} ω}{2}

Potential difference between points O and A is

\frac{B_{0} R^{2} ω}{4}

Electric Potential of point A is more than point B.

Potential difference between points A and B is

B_{0} R^{2} ω

answer is B, C, D.

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Detailed Solution

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$E_{OA} = \frac{1}{2} B ω R^{2}$

$OA = R$

$E_{OB} = \frac{3}{2} B_{0} ω R^{2}$

$OB = \sqrt{R^{2} + R^{2} + 2 R^{2} \cos 120 °}$

$= \sqrt{3} R$

$V_{A} > V_{B};$

$V_{A} - V_{B} = B_{0} R^{2} ω$

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