A 40-turn circular coil (radius = 4.0 cm, total resistance = 0.20 ohm) is placed

Question

A 40-turn circular coil (radius = 4.0 cm, total resistance = 0.20 ohm) is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as given by B = 50 sin (10pi t) mT where t is measured in seconds. What is the magnitude of the induced current in the coil at 0.10 s? A rod (length = 10 cm) moves on two horizontal frictionless conducting rails. The magnetic field in the region is directed perpendicularly to the plane of the rails and is uniform and constant. If a constant force of 0.60 N moves the bar at a constant velocity of 2.0 m/s, what is the current through the 12 ohm load resistor?

Explanation / Answer

15) N = 40 , r = 4 cm

R = 0.2 ohms , t = 0.1 s

B = 50 sin(10pi*t) mT

dB/dt = 50*10*pi*10^-3* cos(10pi*t)

induced emf = -NAdB/dt

= -40*3.14*0.04^2*50*10*3.14*10^-3*cos(10*3.14*0.1)

V = IR

-40*3.14*0.04^2*50*10*3.14*10^-3*cos(10*3.14*0.1)= I*0.2

I = 1.58 A

16) v = 2 m/s , L = 10 cm , F = 0.6 N, R = 12 ohms

V = IR

V = I*12

induced emf V = BvL

I*12 = B*2*0.1 ..(1)

Magnetic force F = ILB

0.6 = I*0.1*B

B = 6/I

from (1)

I*12 = 6*0.2/I

I = 0.32 A

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