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Question

A small coil is introduced between the poles of an electro-magnet so that its axis coincides with the magnetic field direction. The cross-sectional area of the coil is equal to S=3.0 mm2, the number of turns in N=60. When the coil turns through 180 about its diameter, a ballistic galvanometer connected to the coil indicates a charge q=4.5μC flowing through it. Find the magnetic induction magnitude between the poles provided the total resistance of the electric circuit equals R=40Ω.

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Solution

Δq=2NBSR
B=(Δq)R2NS
=(4.5×106)(40)(2)(60)(3×106)=0.5T

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