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A conducting rod of length 2 m is placed...

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)`.

Text Solution

Verified by Experts

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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