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Question

Physics Question on Electromagnetic induction

A coil of mean area 500cm2500\, cm^{2} and having 10001000 turns is held perpendicular to a uniform field of 0.40.4 gauss The coil is turned through 180180^{\circ} in 110\frac{1}{10} second. The average induced emf is

A

0.02V0.02\, V

B

0.04V0.04\, V

C

1.4V1.4\, V

D

0.08V0.08\, V

Answer

0.04V0.04\, V

Explanation

Solution

Here, A=500cm2=500×104m2A=500\, cm^{2} = 500 \times 10^{-4}\, m^{2} N=1000,B=0.4G=0.4×104TN=1000, B=0.4\,G=0.4 \times 10^{-4}\,T t=110st=\frac{1}{10}\,s When the coil is held perpendicular to the field, the normal to the plane of the coil makes an angle of 00^{\circ} with the field BB. \therefore initial flux, ϕ1=BAcos0=BA\phi_{1}=BA\, cos\, 0^{\circ}=BA final flux, ϕ2=BAcos180=BA\phi_{2}=BA\, cos\, 180^{\circ}=-BA ε=N((ϕ2ϕ1)t)=N(BABAt)\varepsilon =-N (\frac{(\phi_{2}-\phi_{1})}{t})=-N (\frac{-BA-BA}{t}) =2NBAt=\frac{2NBA}{t} =2×1000×0.4×104×500×1041/10=\frac{ 2\times1000\times0.4\times10^-4\times500\times10^{-4} }{ 1/10} =0.04V=0.04 \, V