A coil of 40 $Ω$ resistance has 100 turns and radius 6 mm is connected to ammeter of resistance of 160 ohms. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, 32 $μ$ C charge flows through it. The intensity of magnetic field will be

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5.66 T

0.655 T

0.566 T

6.55 T

answer is D.

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Detailed Solution

$\begin{array}{l} q = - \frac{N}{R} (B_{2} - B_{1}) Acos θ \\ 32 \times 10^{- 6} = - \frac{100}{(160 + 40)} (0 - B) \times π \times {(6 \times 10^{- 3})}^{2} \times \cos 0^{\circ} \\ \Rightarrow B = 0 .565 T \end{array}$

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