Question

The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO. Calculate (a) the flux through the loop as a function of time, (b) the emf induced in the loop, (c) the current induced in the loop for a loop resistance of 1.00 V, (d) the power delivered to the loop, and (e) the torque that must be exerted to rotate the loop.

Answer 1

Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

Emf=-NdΦ/dt

Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

i=-0.96NCos(120πt) Ampere

d. Power delivered to the loop

Power is given as

P=IV

P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm