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  • A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Answers (1)

 The long straight wire carrying 25 A current resting on the table applies a force on PQ. And if two straights wires are placed parallel with the current in opposite direction, the forces are repulsive. But if there is an equilibrium in wire PQ, then the repulsive force on PQ balances its weight

The magnetic field produced by a long straight wire carrying current of 25A rests on the table on small wire.

B= \frac{\mu _{0}I}{2 \pi h}

The magnetic force on the small conductors 

F = BlI \sin \theta =BIl

Force applied on PQ balance the weight of the small current carrying capacity

F=mg= \frac{\mu _{0}I^{2}l}{2 \pi h}

h= \frac{\mu _{0}I^{2}l}{2 \pi mg}=\frac{4\pi \times 10^{-7}\times (25)^{2}\times l}{2\pi \times 2.5 \times 10^{-3}\times 9.8 }=5l\times 10^{-4}m = 0.51m

 

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