A square wire loop with 0.600 m sides is perpendicular to a uniform magnetic fie
ID: 2260193 • Letter: A
Question
A square wire loop with 0.600 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 0.68 ? and the internal resistance of the battery is 0.55 ?. If the total emf in the loop is 34.566 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 potential drop across the internal resistance of the battery?
What is the current in the circuit? A square wire loop with 0.600 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 0.68 ? and the internal resistance of the battery is 0.55 ?. If the total emf in the loop is 34.566 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.
Explanation / Answer
a= 0.6^2/2=0.18
db/dt = -8.7
e=-8.7*0.18 =-1.566
emf of battery = 34.566-1.566=33V
I=E/R=34.566/1.23
=28.1
potential drop across the internal resistance = I*0.55
=28.1*0.55
=15.45
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