Question

A coil of 1200 turns and mean area of $ 500\text{ }c{{m}^{2}} $ is held perpendicular to a uniform magnetic field of induction $ 4\times {{10}^{-4}}. $ The resistance of the coil is 20 $ \Omega $ . When the coil is rotated through $ 180{}^\circ $ in the magnetic field in 0.1 s the average electric current (in mA) induced is :

• A. 12
• B. 24
• C. 36
• D. 48

Answer 1

Answer: (B)

In perpendicular position in uniform magnetic field, magnetic field linked with the each turn of coil $ \text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=BA $ On rotating coil through $ {{180}^{o}}, $ flux linked $ \text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}=-BA $ $ \therefore $ Change in magnetic flux $ \Delta \text{o }\!\!|\!\!\text{ }=\text{o}{{\text{ }\!\!|\!\!\text{ }}_{2}}-\text{o}{{\text{ }\!\!|\!\!\text{ }}_{1}}=-2BA. $ $ \therefore $ emf induced in the coil $ e=-N\frac{\Delta \text{o }\!\!|\!\!\text{ }}{\Delta t} $ $ =\frac{2NBA}{\Delta f} $ where N is the number of turns in the coil. The current in the coil $ i=\frac{e}{R}=\frac{2NBA}{R\Delta \Tau } $ $ =\frac{2\times 1200\times 4\times {{10}^{-4}}\times 500\times {{10}^{-4}}}{20\times 0.1} $ $ =2400\times {{10}^{-5}} $ $ =24\times {{10}^{-3}}A $ $ =24\,mA $