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The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through each turn of the coil is

a
14π μ0Wb
b
12πμ0Wb
c
13πμ0Wb
d
0.4 μ0Wb

detailed solution

Correct option is A

Nϕ=Li⇒ϕ=LiN=8×10−3×5×10−3400=10−7=μ04πWb

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Two different coils have self-inductance  L1=8mH, L2=2mH . The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are i1,V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2,V2  and W2 respectively. Then


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