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Question

A circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of 0.1T normal to the plane of the coil. If the current in the coil is 5.0A what is the average force, on each electron in the coil due to the magnetic filed (The coil is made of copper wire of cross-sectional area 105m2 and the free electron density in copper is given to be about 1029m3)

A
5×1025N
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B
2.5×1025N
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C
7.5×1025N
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D
1025N
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Solution

The correct option is A 5×1025N
Number density of electrons N={1029m3}
Area of cross -section of copper wire, A={105m2} magnitude of magnetic force
F=e({vd×B}xB)=neA{V}{d}{V}{d}=IneAF=e.INeAF.Bsin900=0.1×5105×1029N=5×1025N

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