4. A coil of 50 turns sits in a vertical magnetic field with a variable magnetic

Question

4. A coil of 50 turns sits in a vertical magnetic field with a variable magnetic field. The field varies as a sinusoidal according to the equation B = 5.00sin[(pi/2)t] (where t is in seconds). Each layer of the coil is 0.005 mm and the total coil has a volume of 6.25x10^-4 pi m^3. If the coil has an internal resistance of 5 ohm determine a) the induced voltage in the coil between the times ti = 0.33 s to tf =0.67 s (make sure your calculator is in radian mode!!!) b) the induced current in the coil c). Now suppose the coil oscillates back and forth. Suppose that at ti= 0.33 s the area of the coil is pointed at an angle of 30 degrees from the vertical, and at tf = 0.67 s the coil is 10 degrees from the vertical. Find both quantities from above.

Explanation / Answer

given B = 5 sin (pit/2)

so dB/dt = 7.85 cos ( pi t/2)

so induced emf is defiend as the rate of change of magnetic flux

in mathematical form,

induced emf e = NAdB/dt    or NABW   or   NA dB/dt cos theta

where A = area

N = no. of tunrs

W = angular frequency

dB/dt = rate of change of magnetic field

induced emf e   = 50 * (6.25 e-4 * 3.14/0.005 e-3) * 7.95 cos ( 3.14*0.33/2)

emf e = 1.56 e 5 Volts

so

induced current i = e/R

Where R = resisatnce

i = 1.56 e 5/5

i = 3.12 e 4 Amps

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