Question

By a change of current from 5A to 10 A in 0.1s, the self-induced emf is 10V. The change in the energy of the magnetic field of a coil will be
(A) $5J$
(B) $6J$
(C) $7.5J$
(D) $9J$

Answer 1

Hint: In electromagnetism and hardware, inductance is the inclination of an electrical conduit to contradict an adjustment in the electric flow moving through it. Inductance is characterized as the proportion of the instigated voltage to the pace of progress of flow causing it.

Complete step by step answer:
Given: Initial current flowing in the coil = ${I_1} = 5A$
Final current flowing in the coil = ${I_2} = 10A$
Time required to change current from $5A$ to $10A$= $\Delta t = 0.1s$
The self-induced emf in the coil = $10V$
In electromagnetism and hardware, inductance is the inclination of an electrical conduit to contradict an adjustment in the electric flow moving through it. Inductance is characterized as the proportion of the instigated voltage to the pace of progress of flow causing it.
Hence we can give inductance as :
$|\varepsilon | = L\dfrac{{\Delta I}}{{\Delta t}}$
The difference between initial and final current in the coil is,
\[ L = |\varepsilon |\dfrac{{\Delta t}}{{\Delta I}} = \dfrac{{10 \times 0.1}}{5} = \dfrac{1}{5} = 0.2H\]
The magnetic field energy for initial current and final current flowing through the coil:
${U_1} = \dfrac{1}{2}LI_1^2$
And ${U_2} = \dfrac{1}{2}LI_2^2$
Change in the energy = $\dfrac{1}{2}LI_2^2 - $ ${U_1} = \dfrac{1}{2}LI_1^2$
Change in the energy = $\dfrac{1}{2} \times 0.2 \times {10^2} - \dfrac{1}{2} \times 0.2 \times {5^2}$
Change in the energy = $10 - 2.5$
Change in the energy =$ 7.5J$

Hence option (C) is the correct answer.

Note: As a common sense issue, longer wires have more inductance, and thicker wires have less, similar to their electrical obstruction (in spite of the fact that the connections aren't direct, and are diverse in kind from the connections that length and breadth bear to opposition).