Moving Charges and Magnetism - Result Question 58
59. An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I along the same direction is shown in fig. Magnitude of force per unit length on the middle wire ’ $B$ ’ is given by
[2017]
(a) $\frac{2 \mu_0 i^{2}}{\pi d}$
(b) $\frac{\sqrt{2} \mu_0 i^{2}}{\pi d}$
(c) $\frac{\mu_0 i^{2}}{\sqrt{2} \pi d}$
(d) $\frac{\mu_0 i^{2}}{2 \pi d}$
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Answer:
Correct Answer: 59. (c)
Solution:
- (c) Force per unit length between two parallel current carrying conductors,
$F=\frac{\mu_0 i_1 i_2}{2 \pi d}$
Since same current flowing through both the wires
$i_i=i_2=i$
so $F_1=\frac{\mu_0 i^{2}}{2 \pi d}=F_2$
$\therefore$ Magnitude of force per unit length on the middle wire ‘B’
$F _{\text{net }}=\sqrt{F_1^{2}+F_2^{2}}=\frac{\mu_0 i^{2}}{\sqrt{2} \pi d}$
