AB is a resistanceless conducting rod which forms a diameter of a conducting ring of radius r rotating in a uniform magnetic field B as shown in figure. The resistors R₁ and R₂ do not rotate. Then the current through the resistor R1 is

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$\frac{{Bωr}^{2}}{2 R_{1}}$

$\frac{{Bωr}^{2}}{2 (R_{2} + R_{2})}$

$\frac{{Bωr}^{2}}{2 R_{1} R_{2}} (R_{1} + R_{2})$

$\frac{B ω r^{2}}{2 R_{2}}$

answer is A.

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Detailed Solution

The equivalent diagram is

The induced emf across the centre and any point on the circumference is
$| \vec{e} | = \frac{1}{2} {Bωl}^{2} = \frac{{Bωr}^{2}}{2}$
$∴ Current through R_{1} = \frac{{Bωr}^{2}}{2 R_{1}}$

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