A 1000-turn loop (radius = 0.037 m of wire is connected to a 20.0 ohm resistor a

Question

A 1000-turn loop (radius = 0.037 m of wire is connected to a 20.0 ohm resistor as shown in the figure. A magnetic field is directed perpendicular to the plane of the loop. The field points into the paper and has a magnitude that varies in time as B=gt, where g=0.25 Tesla per second . Neglect the resistance of the wire.

1)What is the magnitude of the potential difference between points a and b?

2)What is the electrical energy dissipated in the resistor in 35.0 s?

A 1000-turn loop (radius = 0.037 m of wire is connected to a 20.0 ohm resistor as shown in the figure. A magnetic field is directed perpendicular to the plane of the loop. The field points into the paper and has a magnitude that varies in time as B=gt, where g=0.25 Tesla per second . Neglect the resistance of the wire. 1)What is the magnitude of the potential difference between points a and b? 2)What is the electrical energy dissipated in the resistor in 35.0 s?

Explanation / Answer

Given

Number of loops N = 1000

Radius of the circle r = 0.037 m

Resistance of the wire R = 20.0 ohm

The instantaneous induction B = gt

Solution

1)

Area of cross section of the loop

A = r2

A = 3.14 x (0.037)2

A = 4.299 x 10-3 m2

Rate of change of magnetic field

dB/dt = d(gt)/dt

dB/dt = g

dB/dt = 0.25 T/s

emf induced in a loop by changing magnetic field

e = -NA(dB/dt)cos

here is the angle between area vector and the direction of magnetic field

The area vector is perpendicular to the area of the loop and directed out of the paper

The magnetic field is perpendicular to the area of the loop and directed into the paper

so = 180o

e = - 1000 x 4.299 x 10-3 x 0.25 x cos108

e = 1.075 V

2)

Power dissipated in the resistor

P = e2/R

P = 1.0752 / 20

P = 57.78 x 10-3 J/s

Energy Dissipated

E = Pt

E = 57.78 x 10-3 x 35

E = 2.02 Joules

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