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Question: A solenoid has 2000 turns wound over a length of 0.3m. The area of its cross section is \(1.2\times ...

A solenoid has 2000 turns wound over a length of 0.3m. The area of its cross section is 1.2×10−3m21.2\times {{10}^{-3}}{{\text{m}}^{\text{2}}}. Around its cross section a coil of 300 is wound. If an initial current of 2A is reversed in 0.25 seconds, the e.m.f induced in the coil is equal to:
A. 6×10−4V6\times {{10}^{-4}}\text{V}
B. 4.8×10−2V4.8\times {{10}^{-2}}\text{V}
C. 2.4×10−2V2.4\times {{10}^{-2}}\text{V}
D. 48KV48\text{KV}

Explanation

Solution

Hint: The change in magnetic flux associated with a coil always induces an e.m.f in the coil nearby to it. This is called mutual induction.

Complete step by step answer:
We know that due to mutual induction the change in magnetic flux or change in current during a particular period in one coil can induce an e.m.f in another coil which is in proximity with the first coil.
The e.m.f induced in the second due to the first coil is given by,
E=−N2A2(dB2dt)\text{E}=-{{\text{N}}_{\text{2}}}{{\text{A}}_{2}}\left( \dfrac{d{{B}_{2}}}{dt} \right)
Where,
A2{{\text{A}}_{2}} is the cross sectional area of the second coil
N2{{\text{N}}_{\text{2}}} is the number of turns in the second coil
The above equation can also be written as,
E=−M(dI2dt)\text{E}=-\text{M}\left( \dfrac{d{{I}_{2}}}{dt} \right)….. equation(1)
Where M is the mutual inductance of the coil.
The mutual inductance between two coils are given by the formula,
M=L1L2\text{M}=\sqrt{{{\text{L}}_{\text{1}}}{{\text{L}}_{\text{2}}}}
Where,
L1{{\text{L}}_{\text{1}}} is the inductance of the first coil
L2{{\text{L}}_{2}} is the inductance of the second coil
The inductance of the coil is given by the formula,
L= !!μ!! 0N2Al\text{L}=\dfrac{{{\text{ }\\!\\!\mu\\!\\!\text{ }}_{\text{0}}}{{\text{N}}^{2}}\text{A}}{\text{l}}
Since both the coils have same surface area A and length l, we can write mutual inductance M as
M= !!μ!! 0N1N2Al\text{M}=\dfrac{{{\text{ }\\!\\!\mu\\!\\!\text{ }}_{\text{0}}}{{\text{N}}_{1}}{{\text{N}}_{2}}\text{A}}{\text{l}}
Therefore, e.m.f induced can be written as,
E=−( !!μ!! 0N1N2Al)(dI2dt)\text{E}=-\left( \dfrac{{{\text{ }\\!\\!\mu\\!\\!\text{ }}_{\text{0}}}{{\text{N}}_{1}}{{\text{N}}_{2}}\text{A}}{\text{l}} \right)\left( \dfrac{d{{I}_{2}}}{dt} \right)
Substituting the values given in the problem into the above equation gives,
E=−(4π×10−7×300×2000×1.2×10−30.3)(2−(−2))0.25E=-\left( \dfrac{4\pi \times {{10}^{-7}}\times 300\times 2000\times 1.2\times {{10}^{-3}}}{0.3} \right)\dfrac{\left( 2-(-2) \right)}{0.25}
E=0.0482V\text{E}=0.0482\text{V}
Therefore the E.M.F induced is 4.8×10−2 V4.8\times {{10}^{-2}}\text{ V}.
So the answer to the question is option (B)- 4.8×10−2 V4.8\times {{10}^{-2}}\text{ V}

Note:
Mutual Inductance is the basic operating principle of the transformer, motors, generators and any other electrical component that interacts with another magnetic field. Then we can define mutual induction as the current flowing in one coil that induces a voltage in an adjacent coil.
The amount of mutual inductance that links one coil to another depends very much on the relative positioning of the two coils. If one coil is positioned next to the other coil so that their physical distance apart is small, then nearly all of the magnetic flux generated by the first coil will interact with the coil turns of the second coil inducing a relatively large e.m.f and therefore producing a large mutual inductance value.