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Two long, straight wires, each carrying an electric current of 5.0 A are kept parallel to each other at a separation of 2.5 cm. Find the magnitude of the magnetic force experienced by 10 cm of a wire

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a

1 .0 \times 10^{- 5} N

b

3 .0 \times 10^{- 5} N

c

2 .0 \times 10^{- 5} N

d

6 .0 \times 10^{- 5} N

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detailed solution

Correct option is C

The field at the site of one wire due to the other is $B = \frac{μ_{0} i}{2 πd} = \frac{(2 \times 10^{- 7} {TmA}^{- 1}) (5 .0 A)}{2 .5 \times 10^{- 2} m} = 4 .0 \times 10^{- 5} T$
The force experienced by 10 cm of this wire due to the other is
$\begin{array}{l} F = ilB \\ = (5 .0 A) (10 \times 10^{- 2} m) (4 .0 \times 10^{- 5} T) = 2 .0 \times 10^{- 5} N . \end{array}$

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