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NCERT Class XII Chapter
Electromagnetic Induction
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Question : 3 of 17
Marks: +1, -0
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2\mathrm{cm}^2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Solution:  
When the current changes through the solenoid, a change in magnetic field also take place within it.
Initial magnetic field in solenoid,
B1B_1 = μ0nI1\mu_0 n I_1 = 4π × 10−710^{-7} × 1510−2\frac{15}{10^{-2}} × 2 = 120π × 10−510^{-5} T
Final magnetic field , B2B_2 = μ0nI2\mu_0 n I_2
B2B_2 = 4π × 10−710^{-7} × 1510−2\frac{15}{10^{-2}} × 4 = 240π × 10−510^{-5} T
Initial flux through coil inside solenoid placed normal to axis.
ϕi\phi_i = B1AB_1A = 120π × 10−510^{-5} × 2 × 10−410^{-4}
ϕi\phi_i = 240π × 10−910^{-9} Wb
Final flux, ϕf\phi_f = B2AB_2A = 240π × 10−510^{-5} × 2 × 10−410^{-4}
ϕf\phi_f = 480π × 10−910^{-9} Wb
Induced emf
ε = −(ϕf−ϕi)t\frac{-(\phi_f - \phi_i)}{t} = −240×10−9×3.140.1\frac{-240 \times 10^{-9} \times 3.14}{0.1} = - 7.5 µV
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