A square wire loop with 2.000 m sides is perpendicular to a uniform magnetic fie

Question

A square wire loop with 2.000 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 0.55 ? and the internal resistance of the battery is 0.20 ?. If the total emf in the loop is 89.400 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 ,

B = 0.042 - 8.7t

dB/dt = -8.7 T/s

a = 2 m

R = 0.55 Ohm

Ri = 0.2 Ohm

induced emf generated , E= 8.7 * (2^2)/2

induced emf generated , E = 17.4 V

Now ,

for the emf of the battery V

V + E = 89.4

V + 17.4 = 89.4

V = 72 V

the emf of the battery is 72 V

--------------------------

Using Ohm's law

current in the circuit = 89.4/(0.55 + 0.20)

current in the circuit = 119.2 A

-------------------

potential drop across the internal resistance = current * Ri
potential drop across the internal resistance = 119.2 * 0.20
potential drop across the internal resistance = 23.84 V

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