Current Electricity and Electrical InstrumentHard

Question

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current $I = 10sin(\omega t)A$, where $\omega = 1000rad./s$. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed $v = 1\text{ }cm/s$. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is $\alpha/\sqrt{2}\mu\text{ }A$. The value of $\alpha$ is

[Resistance of the loop $= 10\Omega$ ]

Options

A.197
B.80
C.280
D.100

Solution

EMF induced $\varepsilon = A\frac{dB}{dt} = A\mu_{0}n\frac{di}{dt}\varepsilon = A\mu_{0}ni_{0}\omega cos\omega t$ current induced $i = \frac{\varepsilon}{R} = \frac{\pi r^{2}\mu_{0}{ni}_{0}\omega}{R}cos\omega t$ So $i = \frac{\pi r^{2}\mu_{0}ni_{0}\omega}{\sqrt{2}R}$

$${= \frac{\pi \times 10^{- 4} \times 4\pi \times 10^{- 7} \times 500 \times 10 \times 10^{3}}{\sqrt{2} \times 10} }{= \frac{20\pi^{2}}{\sqrt{2}} \times 10^{- 6} }{\simeq \frac{197}{\sqrt{2}}\mu\text{ }A}$$

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