Question

A circular coil of radius 8 cm, 400 turns and resistance 2$\Omega$ is placed with its plane perpendicular to the horizontal component of the earths magnetic field. It is rotated about its vertical diameter through ${180}^o$ in 0.30 s Horizontal component of earth magnetic field at the place is $3x{10}^{-5}$ T. The magnitude of current induced in the coil is approximately

• A. $4\times {10}^{-2}A$
• B. $8\times {10}^{-4}A$
• C. $8\times {10}^{-2}A$
• D. $1.92\times {10}^{-3}A$

Answer 1

The correct option is B $8\times {10}^{-4}A$

Initial flux through the coil

${\varphi }_i=BAcos\theta$

$=3\times {10}^{-5}\times \pi \times (8\times {10}^{-2}{)}^2\times cos{0}^o$

$=192\pi \times {10}^{-9}Eb$

Final flux after the roatation

${\varphi }_f=3\times {10}^{-5}\pi \times (8\times {10}^{-2}{)}^2\times {180}^o$

$=-192\pi \times {10}^{-9}Wb$

The magnitude of induced emf is:

$ϵ=N\frac{|d\varphi |}{dt}=\frac{N|{\varphi }_f={\varphi }_i|}{dt}$

$=\frac{400\times (380\pi \times (10{)}^{-9}}{0.30}=1.6\times {10}^{-3}(-3)$

$I=\frac{ϵ}{R}=\frac{1.6\times {10}^{-3}}{2}=8\times {10}^{-4}A$