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A conducting rod of length 2 m is placed on a horizontal table in north-south direction. It carries a current of 5A from south to north. Find the direction and magnitude of the magnetic force acting on the rod. Given that the Earth's magnetic field at the place is 0.6xx10^(-4)T and angle of dip is (pi)/(6). |
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Answer» Solution :Here LENGTH of rod `l=2m`, current flowing in rod I = 5A, Earth.s MAGNETIC field `B_(E )=0.6xx10^(-4)T` and angle of dip `delta=(PI)/(6)`. As rod is placed horizontaly in north-south DIRECTION and current is flowing south to north, there is no magnetic force due to `B_(H)`, the horizontal component of earth.s field and force is solely due to VERTICAL component `B_(V)` of the earth.s magnetic field which is acting vertically downward. `therefore F=I l B_(V)=I l B_(E ) sin delta=5xx2xx(0.6xx10^(-4))xx"sin"(pi)/(6)=3xx10^(-4)N` In accordance with Fleming.s left hand rule the force F is in horizontal plane directed from east to west direction. |
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