Question

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.

Hint:

For transformers, use the transformation ratio \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Efficiency relates input and output power.

Answer 1

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: \[ F = I_C B l = 5 \times 2.67 \times 10^{-5} \times 0.15 = 2.0 \times 10^{-5} \, \mathrm{N}. \] --- 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: \[ \text{Input Power} = \frac{8000}{0.9} = 8888.89 \, \mathrm{W}. \]