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Question: What will be the energy stored in the magnetic field when the current of \(5A\) produces the magneti...

What will be the energy stored in the magnetic field when the current of 5A5A produces the magnetic flux of 2×103Wb2 \times {10^{ - 3}}Wb through a coil of 500 turns?
(A) 2.5J2.5J
(B) 0.25J0.25J
(C) 250J250J
(D) None of the above

Explanation

Solution

Hint Use the Faraday’s second law of electromagnetic induction and use the formula of self – induction to determine the induced emf. Use both the equation and find the formula for inductance. Then, use the formula of energy stored in the conductor.
Formula used: - The energy stored in the conductor in the terms of inductance can be expressed as –
E=12LI2E = \dfrac{1}{2}L{I^2}
where, LL is the inductance, and
II is the current passing through the conductor

Complete Step by Step Solution
Magnetic flux can be defined as the number of magnetic field lines passing perpendicular to the surface. In this question, there is self – inductance so it can be defined as induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit.
Now, using the Faraday’s second law of electromagnetic induction, we get –
e=dϕdte = - \dfrac{{d\phi }}{{dt}}
If number of turns is equal to NN
e=Ndϕdt(1)\therefore e = N\dfrac{{d\phi }}{{dt}} \cdots \left( 1 \right)
We know that –
e=LdIdt(2)e = L\dfrac{{dI}}{{dt}} \cdots \left( 2 \right)
From equation (1)\left( 1 \right) and (2)\left( 2 \right), we get –
Ndϕdt=LdIdt Nϕ=LI L=NϕI  \therefore N\dfrac{{d\phi }}{{dt}} = L\dfrac{{dI}}{{dt}} \\\ \Rightarrow N\phi = LI \\\ \therefore L = \dfrac{{N\phi }}{I} \\\
Now, according to the question, it is given that,
I=5AI = 5AMagnetic field current,
ϕ=2×103Wb\phi = 2 \times {10^{ - 3}}WbMagnetic flux,
Number of turns in the coil, N=500N = 500
E=12LI2E = \dfrac{1}{2}L{I^2}Now, calculating the energy stored in the capacitor,
LLPutting the value of in the above formula, we get –
E=12(NϕI)I2 E=12NϕI  E = \dfrac{1}{2}\left( {\dfrac{{N\phi }}{I}} \right){I^2} \\\ \Rightarrow E = \dfrac{1}{2}N\phi I \\\
Putting the value of number of turns, magnetic flux and current in the above formula, we get –
E=12×500×2×103×5E = \dfrac{1}{2} \times 500 \times 2 \times {10^{ - 3}} \times 5
Doing the calculations, we get –
E=2.5J\Rightarrow E = 2.5J

Hence, the correct option is (A).

Note: There are two laws of electromagnetic induction given by Faraday –
1. Whenever the magnetic flux passing through the surface changes with respect to time there will be an induced emf generated.
2. Induced emf is always equal to the negative of rate of change of magnetic flux.