Question

# A long horizontally fixed wire carries a current of $$100$$ ampere. Directly above and parallel to it is a fine wire that carries a current of $$20$$ ampere and weights $$0.04$$ newton per meter. The distance between the two wires for which the upper wire is just supported by magnetic repulsion is:

A
102mm
B
102cm
C
102m
D
102km

Solution

## The correct option is C $$10^{-2}m$$Force on the wire element per unit length is given by:$$F = \cfrac{\mu_o I_1 I_2}{ 2 \pi r}$$This balances the weight of the wire per unit length$$0.04 = \cfrac{2 \times 10^{-7} \times 100 \times 20}{r}$$$$r = 10^{-2}\ m$$Physics

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