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Question
A circular coil of radius 2.00 cm has 50 turns. A uniform magnetic field B = 0.200 T exists in the space in a direction parallel to the axis of the loop. The coil is now rotated about a diameter through an angle of 60.0°. The operation takes 0.100 s. (a) Find the average emf induced in the coil. (b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω, calculate the net charge crossing a cross-section of the wire of the coil.
Solution
Given:-
Number of turns of the coil, N = 50
Magnetic field through the circular coil,
\[\overrightarrow B=0.200 \text{ T}\]
Radius of the circular coil, r = 2.00 cm = 0.02 m
Angle through which the coil is rotated, θ = 60°
Time taken to rotate the coil, t = 0.100 s
(a) The emf induced in the coil is given by
\[e = - \frac{N ∆ \phi}{∆ t} = \frac{N( \overrightarrow{B_f} . \overrightarrow{A}_f - \overrightarrow{B_i} . \overrightarrow{A}_i )}{T}\]
\[ = \frac{NB . A (\cos 0^\circ - \cos 60^\circ)}{T}\]
\[ = \frac{50 \times 2 \times {10}^{- 1} \times \pi(0 . 02 )^2}{2 \times 0 . 1}\]
\[ = 5 \times 4 \times {10}^{- 5} \times \pi\]
\[ = 2\pi \times {10}^{- 2} V = 6 . 28 \times {10}^{- 3} V\]
(b) The current in the coil is given by
\[i = \frac{e}{R} = \frac{6 . 28 \times {10}^{- 3}}{4}\]
\[ = 1 . 57 \times {10}^{- 3} A\]
The net charge passing through the cross section of the wire is given by
\[Q = it = 1 . 57 \times {10}^{- 3} \times {10}^{- 1} \]
\[ = 1 . 57 \times {10}^{- 4} C\]
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