Two long straight wires, each carrying an electric current of 5 A, are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experiment by 10 cm of a wire.

4.0 × 10^-4 N

3.5 × 10^-6 N

2.0 × 10^-5 N

2.0 × 10^-9 N

Solution

2.0 × 10^-5 N

The field at the site of one wire due to the other is

B = $\frac{μ_{o} i}{2 πd}$

Where μ_o - permeability of free space

i - current

B = $\frac{2 \times 10^{7} \times 5.0}{2.5 \times 10^{- 2}}$

B = 4.0 × 10^-5 T

The force experienced by 10 cm of this wire due to other is

F = il B

= (5.0 ) × ( 10 × 10^-2 ) × ( 4.0 × 10^-5)

F = 2.0 × 10^-5 N

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