) My Notes Ask Your 1 112 points 1 Previaus Anowers SerPSE9 31.P 042 A 105-turn

Question

) My Notes Ask Your 1 112 points 1 Previaus Anowers SerPSE9 31.P 042 A 105-turn square coil of side 20.0 cm rotates about a vertical axis at a- 1.39 x 10 rev/min as indicated in the figure below. The horizontal component of Earth's magnetic field at the coil's location is equal to 2.00 × 10-5 T. 20.0 20.0 cn (a) Calculate the maximum emf induced in the coil by this field. mV (b) What is the orientation of the coil with respect to the magnetic field when the maximum emf occurs? The plane of the coil is paralial to the magnetic field. .

Explanation / Answer

emf = w*NBAsin(wt)

for max emf sin(wt) = 1

emf max = w*NBA = 1390*2pi/60 rad/sec * 105*2*10^-5 * ( 0.20)^2

= 0.0122271 V

= 12.22 mV

this will max when the magnetic field in perpendicular coil plane or

the magnetic field is parallel to the normal or the area vector of coil.

emf = N*d(phi) /dt = N*(BA) dt = NA* dB/dt = NA* d(0.01t +0.04t^2)/dt

= NA * ( 0.01+0.08t)|t=5.20

= NA *( 0.01+0.08*5.20) = NA*0.426

= 29 *pi*(0.042)^2 * 0.426 = 0.0685 V

   = 68.5 mV

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