Problem

Two identical coils $A$ and $B$ of 500 turns each are placed on parallel planes such that $70\%$ of the flux produced by one coil links with the other. A current of 6 A flowing in coil $A$ produces a flux of $0.06 \, \mathrm{mWb}$ in it. If the current in coil $A$ changes from $10 \, \mathrm{A}$ to $-10 \, \mathrm{A}$ in $0.03 \, \mathrm{s}$, calculate:
(a) The mutual inductance.
(b) The e.m.f. induced in coil $B$.

Data

• Number of turns: $N = 500$
• Primary magnetic flux: $\Phi_1 = 0.06 \, \mathrm{mWb} = 0.06 \times 10^{-3} \, \mathrm{Wb}$
• Secondary magnetic flux: $\Phi_2 = 70\% \Phi_1 = 0.7 \times 0.06 \times 10^{-3} \, \mathrm{Wb} = 0.042 \times 10^{-3} \, \mathrm{Wb}$
• Current in primary coil: $I_p = 6 \, \mathrm{A}$
• Initial current: $I_i = 10 \, \mathrm{A}$
• Final current: $I_f = -10 \, \mathrm{A}$
• Time taken for change: $\Delta t = 0.03 \, \mathrm{s}$

To find:

1. Mutual inductance $M$.
2. Induced e.m.f. $\varepsilon_s$ in coil $B$.

Prerequisite Concepts

1. Mutual inductance formula:

$$M = \frac{N \Phi_{BS}}{I_p}$$

where $\Phi_{BS}$ is the magnetic flux linked with the secondary coil. 2. Induced e.m.f. in the secondary coil:

$$\varepsilon_s = -M \frac{\Delta I}{\Delta t}$$

Solution

Step 1: Mutual Inductance

1. Use the formula for mutual inductance:

$$M = \frac{N \Phi_2}{I_p}$$

2. Substituting the given values:

$$M = \frac{500 \times 0.042 \times 10^{-3}}{6}$$

3. Simplify:

$$M = 3.5 \times 10^{-3} \, \mathrm{H} = 3.5 \, \mathrm{mH}$$

Step 2: Induced e.m.f. in Coil B

1. Use the formula for induced e.m.f.:

$$\varepsilon_s = -M \frac{\Delta I}{\Delta t}$$

2. Change in current:

$$\Delta I = I_f - I_i = -10 - 10 = -20 \, \mathrm{A}$$

3. Substituting values:

$$\varepsilon_s = -3.5 \times 10^{-3} \times \frac{-20}{0.03}$$

4. Simplify:

$$\varepsilon_s = 2333.33 \times 10^{-3} \, \mathrm{V} = 2.33 \, \mathrm{V}$$

Answer

1. Mutual inductance:
   $M = 3.5 \, \mathrm{mH}$.
2. Induced e.m.f. in coil $B$:
   $\varepsilon_s = 2.33 \, \mathrm{V}$.

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