A square wire loop with 1.800 m sides is perpendicular to a uniform magnetic fie

Question

A square wire loop with 1.800 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 0.31 ? and the internal resistance of the battery is 0.25 ?. If the total emf in the loop is 38.094 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

Here ,

magnetic field , B = 0.042 - 8.7 t

induced emf , E1 = dB/dt * area

E1 = 8.7 * (1.8^2)/2

E1 = 14.094 V

Now , let the emf of battery is Eb

as Eb + E1 = total emf in loop

14.1 + Eb = 38.094

Eb = 24 V

the emf of the battery is 24 V
-------------------

current in the circuit = total emf/(total resistance)

current in the circuit = 38.094/(0.31 + 0.25)

current in the circuit = 68.03 A

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potential drop across the internal resistance = current * resistance

potential drop across the internal resistance = 68.03 * 0.25

potential drop across the internal resistance = 17 V

the potential drop across the internal resistance is 17 V

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