Question

An electric generator consists of a 10-turn square wire loop of side 50cm. The loop is turned so as to produce 50 Hz A.C. How strong must the magnetic field be for the peak output voltage to be 300 V?
a) 2.4T
b) 0.417T
c)2.62T
d)0.382T​

Answer 1

Answer:

Option : d

Explanation:

Answer 2

The magnetic field should be strong for the peak output voltage to be 300 V is :

• Given : Number of turns in wire = 10,

Side of wire loop = 50 cm, frequency = 50 hz, peak voltage = 300V

• We know that ,

Φ = nBA cosθ and emf = |- dΦ / dt |

• emf = nBA sinθ dθ/dt

dθ/dt = ω and ω = 2πf

emf = nBAω sinθ

• At peak emf value, sinθ = 1

•°• peak emf = nBAω = nBA×2πf

300 = 10×B×50×50×2π×50/100×100

• B = 60 / 50π = 6 / 5π = 0.382 T