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Problem 9: A 90 mm length of wire moves with an upward velocity of 35 m/s between the poles of a magnet. The magnetic field is 80 mT directed to the right. If the resistance in the wire is 5.00 micro ohms, what are the magnitude and direction of the induced current?

Solution

Given    Length of the wire, L = 90 mm = 90 × 10-3 m,          Magnetic field, B = 80 mT = 80 × 10-3 T

               Velocity, v = 35 ms-1,       Resistance, R = 5.00 × 10-3 ohm,  Angle in B and L = 900

Required              Magnitude and direction of current

Formula               ϵ = BLvsinθ,                                       I = ϵ/R

To find the electromotive force, ϵ apply first formula.

ϵ = 80 × 10-3 × 90 × 10-3 × 35 × 1 = 252000 × 10-6 = 2.52 × 10-1 V

Now to find I, use the second formula.

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