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Q.

A 10 cm length of wire with a mass of 20 g is attached frictionlessly to the vertical segments of a wire in which a current I flows. The surrounding has uniform horizontal field B=104G and the direction is shown in figure. What must be the current I (in A) to maintain the 10 cm wire in an equilibrium position?

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answer is 2.

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Detailed Solution

B=104G=1 TFor equilibrium, magnetic force should be equal to weight of wire.IBl=mg⇒I×1×0.10=20×10-3×10⇒I=2 A
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