Solveeit Logo
Login

Question

Question: A solenoid has \(2000\) turns wound over a length of \(0.3\,m\) . The area of its cross section is \...

A solenoid has 20002000 turns wound over a length of 0.3 m0.3\,m . The area of its cross section is 1.2×10−3 m−21.2 \times {10^{ - 3}}\,{m^{ - 2}} . Around its cross section a coil of 300300 turns are wound. If an initial current of 2 A2\,A is reversed in 0.25 s0.25\,s , the emf induced in the coil is equal to
A. 6×10−4 V6 \times {10^{ - 4}}\,V
B. 4.8×10−2 V4.8 \times {10^{ - 2}}\,V
C. 2.4×10−2 V2.4 \times {10^{ - 2}}\,V
D. 48k V48k\,V

Explanation

Solution

It is given that along the central cross section of the solenoid a coil is kept. This coil is placed coaxially so it will experience the magnetic field due to the solenoid. The change in magnetic flux in a coil always induces e.m.f. in the coil to its surrounding. This is known as mutual induction.

Complete step by step answer:
Due to mutual induction the change in magnetic flux or change in electric current during a particular period in one coil can induce emf in another coil.
This induced emf is given by:
E=−N2A2(dB2dt)E = - {N_2}{A_2}\left( {\dfrac{{d{B_2}}}{{dt}}} \right)
Here, A2{A_2} is the cross-sectional area of the second coil
N2{N_2} is the number of turns of the second coil;
The above equation can be modified as:
E=−M(dI2dt)E = - M\left( {\dfrac{{d{I_2}}}{{dt}}} \right)
Here, MM denotes the mutual inductance.
Mutual inductance between two coils is given as:
M=L1L2M = \sqrt {{L_1}{L_2}}
Here, L1{L_1} is inductance of the first coil
L2{L_2} is the inductance of the second coil
The inductance of a coil is given as:
L=μ0N2AIL = \dfrac{{{\mu _0}{N^2}A}}{I}
As both the coils have the same surface area and length, we can write mutual inductance as:
M=μ0N1N2AIM = \dfrac{{{\mu _0}{N_1}{N_2}A}}{I}
Thus, the emf induced can be written as,
E=−(μ0N1N2AI)(dI2dt)E = - \left( {\dfrac{{{\mu _0}{N_1}{N_2}A}}{I}} \right)\left( {\dfrac{{d{I_2}}}{{dt}}} \right)
Substituting the given values, we have
E=−(4π×10−7×300×2000×1.2×10−30.3)(2−(−2)0.25)E = - \left( {\dfrac{{4\pi \times {{10}^{ - 7}} \times 300 \times 2000 \times 1.2 \times {{10}^{ - 3}}}}{{0.3}}} \right)\left( {\dfrac{{2 - ( - 2)}}{{0.25}}} \right)
∴E=0.0482V\therefore E = 0.0482V
The negative sign is neglected. Thus, the emf induced is E=4.8×10−2 VE = 4.8 \times {10^{ - 2}}\,V

Hence,option B is the correct option.

Note: Mutual inductance is the basic operating principle of transformers, motors, generators and other electrical components that interact with magnetic fields. The amount of mutual inductance linked to one coil depends on the relative position of the two coils. When both coils are nearer, a larger emf is generated and when both the coils are far away then the emf induced is relatively less.