SATHEE: Moving Charges and Magnetism

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) $\frac{2 μ_{0} i^{2}}{π d}$

(b) $\frac{\sqrt{2} μ_{0} i^{2}}{π d}$

(d) $\frac{μ_{0} i^{2}}{2 π d}$

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Answer:

Correct Answer: 59. (c)

Solution:

$F = \frac{μ_{0} i_{1} i_{2}}{2 π d}$

Since same current flowing through both the wires

$i_{i} = i_{2} = i$

so $F_{1} = \frac{μ_{0} i^{2}}{2 π d} = F_{2}$

$∴$ Magnitude of force per unit length on the middle wire ‘B’

$F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2}} = \frac{μ_{0} i^{2}}{\sqrt{2} π d}$

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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

Answer:

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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