A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The
current induced in the coil is
(Horizontal component of the earth's magnetic field at that place is 3.0 x 10^-5 T)

1.9 × 10^-3 A

2.9 × 10^-3 A

3.9 × 10^-3 A

4.9 × 10^-3 A

Solution

1.9 × 10^-3 A

Initial magnetic flux through the coil

$ϕ_{i}$ = B_H Acosθ

= 3.0 × 10^-5 × ( $π$ × 10^-2 ) × cos0^o

$ϕ_{i}$ = 3 $π$ × 10^-7 Wb

Final magnetic flux after the rotation

$ϕ_{f}$ = 3.0 × 10^-5 × ( $π$ × 10^-2 )×cos180^o

= 3 $π$ × 10^-7 Wb

lnduced emf,

$ε = - N \frac{dϕ}{dt}$

= $- \frac{N (ϕ_{f} - ϕ_{i})}{t}$

$ε = \frac{500 (- 3 π \times 10^{- 7} - 3 π \times 10^{- 7})}{0.25}$

= $\frac{500 \times (6 π \times 10^{- 7})}{0.25}$

= 3.8 × 10^-3 V

Current is given by

I = $\frac{ε}{R}$

= $\frac{3.8 \times 10^{- 3}}{2 Ω}$

I = 1.9 × 10^-3 A

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