A square wire coil of 100 turns copper wire (diameter 0.2 mm, rho = 1.68 times 1

Question

A square wire coil of 100 turns copper wire (diameter 0.2 mm, rho = 1.68 times 10^-8 Ohm m) and sides of 20.0 cm is located in the xy-plane. The coil is centered at the origin. A l.5 V battery is inserted in the wire loop such that a constant current flows in a clockwise sense, as viewed from above. The magnetic field is given by B(t) = (0.05 - 0.08t)z. B is in units of Tesla and time t in seconds. (a) What is the net EMF induced in the circuit? (b) What is the magnitude of the net current, and in what direction does it flow as viewed from above? (c) What is the power dissipated by the wire?

Explanation / Answer

PART-1: Induced emf is given by,

e = - (d/dt) where, represents magnetic flux

= B A n cos () where, represents the angle between magnetic field and area vector, B represents magnetic field, A represents area and n represents area vector.

Now, = (0.05-0.08t)tesla* ( 20*20*10-4)m2*cos (1800)

= - 4*10-2*(0.05-0.08t) Weber

Now, e = - (d/dt) (- 4*10-2*(0.05-0.08t)) = 4*10-2(d/dt)(0.05-0.08t) = -0.32*10-2 v

PART-2: Induced current is given as, I = e/R where, R represents resistance provided by the loop.

Now, R = (L/A) , where is resistivity, L is length of wire and A=r2 is area of cross section of wire.

R = 1.68*10-8m[(100*4*20*10-2m) / (3.14*10-8m2)

R = 42.8

Now, Ii = e / R = (0.32*10-2v) / (42.8 ) = 7.4 * 10-5 A

Now constant current flowing through the loop I = V / R = (1.5 V) / (42.8 ) = 3.5*10-2 A

Net current Inet = I - Ii = 3.49*10-2 A , this current will flow in clockwise direction as seen from above as the strength of I is greater then Ii.

PART-3: Power dissipated by the wire P = Inet2 * R = (3.49*10-2 A)2*(42.8 ) = 5.2*10-2 Watt

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