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Moving Charges And Magnetism

Question

Wired Faculty App

A wire AB is carrying a current of 12A and is lying on the table. Another wire CD, carrying current 5A is arranged just above AB at a height of 1 mm. What should be the weight per unit length of this wire so that CD remains suspended at its position? Indicate the direction of current in CD and the nature of force between two wires.

Solution

Current on AB, I₁ = 12 A
Current on CD, I₂ = 5 A
Distance between wires AB and CD, d = 1 mm = 10^-3 m

Weight of the wire will be balanced by the force of repulsion between the wires AB and CD so that the wire CD remains suspended at it's position

Therefore,

\frac{F}{l} = \frac{μ_{0} I_{1} I_{2}}{2 πd}

i.e.,

\frac{mg}{l} = \frac{μ_{0} I_{1} I_{2}}{2 πd}

= \frac{4 π \times 10^{- 7} \times 12 \times 5}{2 π \times 1 \times 10^{- 3}} {Nm}^{- 1} = 120 \times 10^{- 4} = 0.12 {Nm}^{- 1}

mg/l is the weight per unit length.

The direction of current in CD will be opposite to that of in AB.

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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