A square wire loop with 1.400 m sides is perpendicular to a uniform magnetic fie

Question

A square wire loop with 1.400 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field, as shown in the figure. The resistance of the wire is 0.29 ohms and the internal resistance of the battery is 0.18 ohms. If the total emf in the loop is 52.526 V calculate the emf of the battery, when the magnitude of the field varies with time according to B = 0.042 - 8.70t, with B in teslas and t in seconds

What is the current in the circuit?

What is the potential drop across the internal resistance of the battery?

A square wire loop with 1.400 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field, as shown in the figure. The resistance of the wire is 0.29 ohms and the internal resistance of the battery is 0.18 ohms. If the total emf in the loop is 52.526 V calculate the emf of the battery, when the magnitude of the field varies with time according to B = 0.042 - 8.70t, with B in teslas and t in seconds What is the current in the circuit? What is the potential drop across the internal resistance of the battery?

Explanation / Answer

Area of the traingle

A=(1/2)L2=(1/2)*1.42=0.98 m2

Induced emf

E=-A(dB/dt) =-0.98*(-8.7)

E=8.526 Volts

so Emf of the battery is

EB=Et-E =52.526-8.526

EB=44 volts

b)

equivalent resistance

R=0.29+0.18=0.47 ohms

Current in the circuit is

I=EB/R =44/0.47

I=93.62 A

c)

Potential drop across the internal resistance is

V=I*r= 93.62*0.18=16.85 volts

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