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 _____. [2026]

Question

Smart Achievers · Accepted Answer

(1)

$F_{net} = \frac{μ_{0}}{2 π} I_{0} (\frac{I_{1}}{d_{1}} + \frac{I_{2}}{d_{2}}) ℓ$

$F_{net} = 2 \times 10^{- 7} \times (\frac{3}{3} + \frac{2}{2}) \times \frac{15 \times 10^{- 2}}{10^{- 2}}$

$= 4 \times 15 \times 10^{- 7}$

$F_{net} = 6 \times 10^{- 6} N$

Solution

Q.
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 _____. [2026]

1 $6 \times 10^{- 6} N towards R$

2 $6 \times 10^{- 7} N towards R$

3 $6 \times 10^{- 7} N towards P$

4 $6 \times 10^{- 6} N towards P$

Ans.
(1)

$F_{net} = \frac{μ_{0}}{2 π} I_{0} (\frac{I_{1}}{d_{1}} + \frac{I_{2}}{d_{2}}) ℓ$

$F_{net} = 2 \times 10^{- 7} \times (\frac{3}{3} + \frac{2}{2}) \times \frac{15 \times 10^{- 2}}{10^{- 2}}$

$= 4 \times 15 \times 10^{- 7}$

$F_{net} = 6 \times 10^{- 6} N$

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Solution

Q. 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 _____. [2026]

1 6×10-6 N towards R

2 6×10-7 N towards R

3 6×10-7 N towards P

4 6×10-6 N towards P

Ans. (1) Fnet=μ02πI0(I1d1+I2d2)ℓ Fnet=2×10-7×(33+22)×15×10-210-2 =4×15×10-7 Fnet=6×10-6 N