Two long conductors $$10$$ cm apart carry currents in the ratio $$1$$ :$$2$$ in the same direction. The magnetic field midway between them is $$2$$ × $$10$$−$$3$$ T The force on unit length of any one conductor will be

Question

Two long conductors $$10$$ cm apart carry currents in the ratio $$1$$ :$$2$$ in the same direction. The magnetic field midway between them is $$2$$ ×  $$10$$−$$3$$ T The force on unit length of any one conductor will be

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Magnetic field at their $$mid-point$$,$$B=$$μ$$02$$$$π$$$$d2$$  $$2i$$−i  $$=$$ μ$$0i$$$$π$$dGiven  μ$$0i$$$$π$$d  $$=$$  $$2$$ ×  $$10$$−$$3$$ TSubstituting d $$=$$ $$0.1$$ m, we get $$i=1032$$ AMagnetic field at the location of second conductor due to first conductor is, $$B=$$μ$$0i2$$$$π$$dForce on unit length of second $$conductor=B$$×  $$2i=$$μ$$0i2$$$$π$$d $$=$$  μ$$0i$$$$π$$d × i $$=$$  $$1N$$

Force on a current carrying wire placed in a magnetic field

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Two long conductors 10 cm apart carry currents in the ratio 1 :2 in the same direction. The magnetic field midway between them is $2 \times 10^{- 3} T$ The force on unit length of any one conductor will be

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Force on a current carrying wire placed in a magnetic field

Remember concepts with our Masterclasses.

Two long conductors 10 cm apart carry currents in the ratio 1 :2 in the same direction. The magnetic field midway between them is 2 × 10−3 T The force on unit length of any one conductor will be

Magnetic field at their mid-point,B=μ02πd2 2i−i = μ0iπdGiven μ0iπd = 2 × 10−3 TSubstituting d = 0.1 m, we get i=1032 AMagnetic field at the location of second conductor due to first conductor is, B=μ0i2πdForce on unit length of second conductor=B× 2i=μ0i2πd = μ0iπd × i = 1N

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Two long conductors 10 cm apart carry currents in the ratio 1 :2 in the same direction. The magnetic field midway between them is $2 \times 10^{- 3} T$ The force on unit length of any one conductor will be