Currents of 10 A and 20 A are passed in two long parallel wires A and B, as shown in the figure. A wire C of 15 cm length is placed just in between these two wires, in which 5 A of current is passed. Then what will be the force acting on the wire C?
(i) Calculate the number of turns in the secondary coil.
(ii)Input power.
Show Hint
For transformers, use the transformation ratio \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Efficiency relates input and output power.
Part 1: Force on wire C
The magnetic field at the position of wire C due to A and B is:
\[
B_A = \frac{\mu_0 I_A}{2\pi d}, \quad B_B = \frac{\mu_0 I_B}{2\pi d}.
\]
Since \( d = 0.075 \, \mathrm{m} \), the net magnetic field \( B \):
\[
B = B_B - B_A = \frac{\mu_0}{2\pi d} (I_B - I_A).
\]
Substituting values:
\[
B = \frac{4\pi \times 10^{-7}}{2\pi \times 0.075} (20 - 10) = 2.67 \times 10^{-5} \, \mathrm{T}.
\]
The force on wire C is:
\[
F = I_C B l = 5 \times 2.67 \times 10^{-5} \times 0.15 = 2.0 \times 10^{-5} \, \mathrm{N}.
\]
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Part 2: Step-down transformer
The transformation ratio is:
\[
\frac{V_s}{V_p} = \frac{N_s}{N_p}.
\]
Given \( V_s = 220 \, \mathrm{V}, V_p = 2200 \, \mathrm{V}, N_p = 5000 \):
\[
N_s = \frac{V_s}{V_p} \times N_p = \frac{220}{2200} \times 5000 = 500.
\]
Efficiency (\( \eta \)):
\[
\eta = \frac{\text{Output Power}}{\text{Input Power}} \quad \Rightarrow \quad \text{Input Power} = \frac{\text{Output Power}}{\eta}.
\]
Substituting values:
\[
\text{Input Power} = \frac{8000}{0.9} = 8888.89 \, \mathrm{W}.
\]
