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Question
A conducting square loop having edges of length 2.0 cm is rotated through 180° about a diagonal in 0.20 s. A magnetic field B exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field.
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Solution
Given:-
Induced emf, e = 20 mV = 20 × 10−3 V
Area of the loop, A = (2 × 10−2)2 = 4 × 10−4 m2
Time taken to rotate the loop, Δt = 0.2 s
The average induced emf is given by
\[e = - \frac{∆ \phi}{∆ t} = \frac{\phi_i - \phi_f}{t}\]
\[\phi = \overrightarrow{B} . \overrightarrow{A} = BA\cos\theta\]
\[ \phi_i = B(4 \times {10}^{- 4} )\cos0 = B(4 \times {10}^{- 4} )\]
\[ \phi_f = B(4 \times {10}^{- 4} )\cos {180}^o = - B(4 \times {10}^{- 4} )\]
\[e = \frac{B(4 \times {10}^{- 4} ) - \left[ - B(4 \times {10}^{- 4} ) \right]}{0 . 2}\]
\[20 \times {10}^{- 3} = \frac{8B \times {10}^{- 4}}{2 \times {10}^{- 1}}\]
\[\Rightarrow 20 \times {10}^{- 3} = 4 \times B \times {10}^{- 3} \]
\[ \Rightarrow B = \frac{20 \times {10}^{- 3}}{4 \times {10}^{- 3}} = 5 T\]
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