Question

A 0.35 m2 coil with 50 turns rotates at 5 radians per sec2 in a magnetic field of 0.6 Tesla. What is the value of the rms current in the coil if the coil has the resistance of 3.3 Ω?

Answer 1

Answer:

11.25 amps

Explanation:

For transformers, the magnetic flux

$\Phi _{max} = \beta \times A$

Therefore;

$\Phi _{max} = 0.6 \times 0.35 = 0.21 \ Weber$

Ф = Фmax (cosωt) = 0.21·(cos(5·t))

From Faraday's law of induction, we have;

ε = -N × dΦ/dt

Which gives;

dΦ/dt = -1.05(sin (5t) )

ε = -N × dΦ/dt = -50× -1.05(sin (5t) )

ε = 52.5(sin (5t) )

I = ε/R = 52.5(sin (5t) )/3.3 = 15.9091(sin (5t) ) amps

The peak current is therefore = 15.9091 amps

The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.