Two horizontal parallel straight conductors, each $$20$$ cm long, are arranged one vertically above the other and carry equal currents in opposite directions. The lower conductor is fixed while the other is free to move in guides remaining parallel to the lower. If the upper conductor weighs $$1.0$$ g, what is the approximate current (in$$ $$A) that will maintain the conductors at a distance $$0.50$$ cm apart?

Question

Infinity Learn · Accepted Answer

In equilibrium magnetic force Fm will balance weight $$mgSomg=Fm$$⇒$$mg=$$μ$$0i2l2$$$$π$$d ⇒  $$i=2$$$$π$$mgdμ$$0l=2$$π×$$2.0$$×$$10-3$$×$$10$$×$$0.50$$×$$10-24$$π×$$10-7$$×$$20$$×$$10-2$$ $$=2500=50$$ A

Magnetic force

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Magnetic force

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In equilibrium magnetic force Fm will balance weight mgSomg=Fm⇒mg=μ0i2l2πd ⇒ i=2πmgdμ0l=2π×2.0×10-3×10×0.50×10-24π×10-7×20×10-2 =2500=50 A

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