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

# 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

1. A

$1.0×{10}^{-5}\mathrm{N}$

2. B

$3.0×{10}^{-5}\mathrm{N}$

3. C

$2.0×{10}^{-5}\mathrm{N}$

4. D

$6.0×{10}^{-5}\mathrm{N}$

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### Solution:

The field at the site of one wire due to the other is   $\mathrm{B}=\frac{{\mathrm{\mu }}_{0}\mathrm{i}}{2\mathrm{\pi d}}=\frac{\left(2×{10}^{-7}{\mathrm{TmA}}^{-1}\right)\left(5.0\mathrm{A}\right)}{2.5×{10}^{-2}\mathrm{m}}=4.0×{10}^{-5}\mathrm{T}$
The force experienced by 10 cm of this wire due to the other is
$\begin{array}{l}\mathrm{F}=\mathrm{ilB}\\ =\left(5.0\mathrm{A}\right)\left(10×{10}^{-2}\mathrm{m}\right)\left(4.0×{10}^{-5}\mathrm{T}\right)=2.0×{10}^{-5}\mathrm{N}.\end{array}$

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