First slide

Force on a current carrying wire placed in a magnetic field

Question

Consider three long, straight, parallel wires shown in fig. The force experienced by 25 cm length of wire C is

Moderate

A

1x 10^-4 N towards right
B

3x 10^-4 N towards left
C

3x 10^-4 N towards right
D

2 x 10^-4 N towards left

Solution

Force on 25 cm long wire C due to A is
$\begin{aligned} F_{CA} = \frac{μ_{0}}{4 π} \times \frac{2 i_{A} \times i_{C}}{3 \times 10^{- 2}} \times \frac{25}{100} towards right \\ = 10^{- 7} \times \frac{2 \times 30 \times 10 \times 25}{3} N = 5 \times 10^{- 4} N \end{aligned}$
Force on 25 cm long wire C due to B is
$\begin{aligned} F_{CB} = \frac{μ_{0}}{4 π} \times \frac{2 i_{B} \times i_{C}}{5 \times 10^{- 2}} \times \frac{25}{100} towards left \\ = 10^{- 7} \times \frac{2 \times 10 \times 20 \times 25}{5} N = 2 \times 10^{- 4} N \end{aligned}$
Net force F = (F_CA - F_CB) towards right
= 3x 10^-4 N towards right

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