First slide

Force on a current carrying wire placed in a magnetic field

Question

Three long, straight and parallel wires carrying current are arranged as shown in figure. The force experienced by 10 cm length of wire Q is

Moderate

A

$1 .4 \times 10^{- 4} N towards right$
B

$1 .4 \times 10^{- 4} N towards left$
C

$2.6 \times 10^{- 4} N towards right$
D

$2.6 \times 10^{- 4} N towards left$

Solution

Force on wire Q due to wire P is
$F_{P} = 10^{- 7} \times \frac{2 \times 30 \times 10}{0 .1} \times 0 .1 = 6 \times 10^{- 5} N$
(Towards Left)
Force on wire Q due to wire R is
$F_{R} = 10^{- 7} \times \frac{2 \times 20 \times 10}{0 .02} \times 0 .1 = 20 \times 10^{- 5} N$
(Towards Right)
$Hence F_{net} = F_{R} - F_{P} = 14 \times 10^{- 5} N = 1 .4 \times 10^{- 4} N$
(Towards Right)

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