Question

Two wires shown in the figure are connected in a series circuit and the same amount of current of 10 A passes through both, but in apposite directions. Separation between the two wires is 8 mm. The length AB is S = 22 cm. Obtain the direction and magnitude of the magnetic field due to current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. [µ₀ = 4π × 10^-7 T.m/A]

Answer 1

Data: I₁ = I₂ = 10 A, s = 8 mm = 8 × 10^-3 m, l = 0.22 m
By right hand grip rule, the direction of the magnetic field \(\overrightarrow{B_{2}}\) due to the current in wire 2 at AB is into the page and its magnitude is
B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{s}=10^{-7} \times \frac{2(10)}{8 \times 10^{-3}}=\frac{1}{4} \times \mathbf{1 0}^{-3}\) T
The current in segment AB is upwards. Then, by Fleming’s left hand rule, the force on it due to \(\overrightarrow{B_{2}}\) is to the left of the diagram, i.e., away from wire 1, or repulsive. The magnitude of the force is
F_{on 1 by 2} = I₁lB₂ = (10)(0.22) × \(\frac{1}{4}\) × 10^-3
= 5.5 × 10^-4 N