First slide

Magnetic flux and induced emf

Question

A long solenoid having 200 turns per centimeter carries a current of 1.5 A. At the center of the solenoid is placed a coil of 100 turns of cross-sectional area 3.14 x 10^-4 m² having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 s, the induced emf in the coil is

Moderate

A

0.48 V
B

0.048 V
C

0.0048 V
D

48 V

Solution

$B = μ_{0} n i = 4 π \times 10^{- 7} \times 200 \times 10^{2} \times 1.5 = 3.8 \times 10^{- 2} T$
$ϕ = B A = 3.8 \times 10^{- 2} \times 3.14 \times 10^{- 4} = 1.2 \times 10^{- 5} Wb$
When the current in the solenoid is reversed, the change in magnetic flux,
$d ϕ = 2 \times (1.2 \times 10^{- 5}) = 2.4 \times 10^{- 5} Wb$
$∴ Induced e.m.f.$
$ε = N (\frac{d ϕ}{d t}) = 100 \times (\frac{2.4 \times 10^{- 5}}{0.05}) = 0.048 V$

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