A long solenoid having 200 turns per cm caries a current of 1.5 amp. At the centre of it 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 sec., the induced e.m.f. in the coil is

B=μ0ni=4π×10−7200×10−2×1⋅5=3⋅8×10−2Wb/m2Magnetic flux through each turn of the coilΦ=BA=3⋅8×10−23⋅14×10−4=1⋅2×10−5 webreWhen the current in the solenoid is reversed, the change in magnetic flux=2×1⋅2×10−5=2⋅4×10−5 weberInduced e.m.f.=NdΦdt=100×2⋅4×10−50⋅05=0⋅048V

A long solenoid having 200 turns per cm caries a current of 1.5 amp. At the centre of it 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 sec., the induced e.m.f. in the coil is

B=μ0ni=4π×10−7200×10−2×1⋅5=3⋅8×10−2Wb/m2Magnetic flux through each turn of the coilΦ=BA=3⋅8×10−23⋅14×10−4=1⋅2×10−5 webreWhen the current in the solenoid is reversed, the change in magnetic flux=2×1⋅2×10−5=2⋅4×10−5 weberInduced e.m.f.=NdΦdt=100×2⋅4×10−50⋅05=0⋅048V

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A long solenoid having 200 turns per cm caries a current of 1.5 amp. At the centre of it is placed a coil of 100 turns of cross-sectional area 3.14 x 10^-4 m²having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 sec., the induced e.m.f. in the coil is

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0.48 V

0.048 V

0.0048 V

48 V

answer is B.

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Detailed Solution

$\begin{array}{l} B = μ_{0} ni \\ = (4 π \times 10^{- 7}) (200 \times 10^{- 2}) \times 1 \cdot 5 = 3 \cdot 8 \times 10^{- 2} Wb / m^{2} \end{array}$
Magnetic flux through each turn of the coil
$\begin{array}{l} Φ = BA = (3 \cdot 8 \times 10^{- 2}) (3 \cdot 14 \times 10^{- 4}) \\ = 1 \cdot 2 \times 10^{- 5} webre \end{array}$
When the current in the solenoid is reversed, the change in magnetic flux
$= 2 \times (1 \cdot 2 \times 10^{- 5}) = 2 \cdot 4 \times 10^{- 5} weber$
Induced e.m.f.
$= N \frac{dΦ}{dt} = 100 \times \frac{2 \cdot 4 \times 10^{- 5}}{0 \cdot 05} = 0 \cdot 048 V$