Question

As shown in the figure, there exists a uniform magnetic field B into the plane of the paper. A ring of radius $\ell$ and total resistance R is fixed. Rod OA and 0B are resistanceless and rotating with constant angular velocity $\omega$ in the same plane as that ring, about hinge point '0' and the separation between O and C (centre of ring) is' $2\ell$ as shown in the figure. Find the current in the rod 'OA' when the both rod becomes tangent to the ring.

• A. Zero
• B. $\frac{BaV}{R}e\frac{-3t}{2CR}$
• C. $\frac{BaV}{R}e\frac{3t}{2CR}$
• D. $\frac{BaV}{R}e\frac{-4t}{2CR}$
• E. None of the above

Answer 1

The correct option is B $\frac{BaV}{R}e\frac{-3t}{2CR}$

Induced emf $\epsilon =VBl\left(\frac{\sqrt{3}}{2}\right)$

$=\frac{\omega }{2}\left(2\sqrt{3}l\right)(\sqrt{3}\times 2l)B=\frac{3}{2}\omega l^{2}B$

applying K V L where $\frac{2R}{3}\parallel \frac{R}{3}$

Requavel Resis .between $R_{eq}=\frac{2R^2}{\frac{9}{R}}=\frac{2R}{9}$

$i=\frac{2e}{R_{eq}}=\frac{2\times \frac{3}{2}\times B\omega l^2}{\frac{2}{9}R}$

$=\frac{27}{2R}B\omega l^2$

At steady state capacitor is short circuited before t= 0 i.e, t<0 emf induced e = 2 aVB downward direction.

In fig (3) current, I through R is zero at t = 0 circuits like.

In fig(4) $\frac{1}{2}=e$

In capacitor number

$V_C=\frac{e}{2}$ sudden voltage change equivalent capacitance = C series 2C $C_{eq}=\frac{2}{3}C$

$I_0=\frac{e-\frac{e}{2}}{R}=\frac{VaB}{R}I=I_{0}e\frac{-3t}{2cR}I=\frac{BaV}{R}e\frac{-3t}{2CR}$