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Question: A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2 T. The normal...

A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2 T. The normal to the plane of the coil makes an angle of 60° with the direction of the magnetic field. The magnetic flux linked with the coil is

A

5×10−3 Wb5 \times 10 ^ { - 3 } \mathrm {~Wb}

B

5×10−5 Wb5 \times 10 ^ { - 5 } \mathrm {~Wb}

C

10−2 Wb10 ^ { - 2 } \mathrm {~Wb}

D

10−4 Wb10 ^ { - 4 } \mathrm {~Wb}

Answer

5×10−3 Wb5 \times 10 ^ { - 3 } \mathrm {~Wb}

Explanation

Solution

ϕ=NBAcos⁡θ=100×0.2×5×10−4cos⁡60∘\phi = N B A \cos \theta = 100 \times 0.2 \times 5 \times 10 ^ { - 4 } \cos 60 ^ { \circ }

=5×10−3 Wb= 5 \times 10 ^ { - 3 } \mathrm {~Wb}