Magnetic field due to currentMedium
Question
Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire $Q$ is $\_\_\_\_$ .
( $\mu_{0} = 4\pi \times 10^{- 7}\text{ }T.m/A$ )
Options
A.$6 \times 10^{- 7}\text{ }N$ towards $P$
B.$6 \times 10^{- 6}\text{ }N$ towards $R$
C.$6 \times 10^{- 7}\text{ }N$ towards $R$
D.$6 \times 10^{- 6}\text{ }N$ towards $P$
Solution
$F_{\text{net~}} = \frac{\mu_{0}}{2\pi}I_{0}\left( \frac{I_{1}}{{\text{ }d}_{1}} + \frac{I_{2}}{{\text{ }d}_{2}} \right)\mathcal{l}$
$${F_{\text{net~}} = 2 \times 10^{- 7} \times 1\left( \frac{3}{3} + \frac{2}{2} \right) \times \frac{15 \times 10^{- 2}}{10^{- 2}} }{= 4 \times 15 \times 10^{- 7} }{F_{\text{net~}} = 6 \times 10^{- 6}\text{ }N}$$
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