Electromagnetism - Faraday’s law

A coil of 15 turns, each of radius 1 centimeter, is rotating at a constant angular velocity ω = 300 radians per second in an uniform magnetic field of 0.5 Tesla, as shown in figure. Assume at time t = 0 that the normal n̂ to the coil plane is along the y-direction and that the self-inductance of the coil can be neglected. If the coil resistance is 9 ohms, what will be the magnitude of the induced current in milliamperes?

A. 225π sin ωt
B. 250π sin ωt
C. 0.08π cos ωt
D. 1.7π cos ωt
E. 25π cos ωt

(GR0177 #86)

Solution:

Ohm’s Law: ɛ = V = IR
→  I = ɛ / R

Faraday's Law: ɛ = − dΦ_B / dt

Magnetic Flux:  dΦ_B =  NB_⊥dA

At time t = 0, the normal to the coil plane is along the y-direction.

It means: n̂ ∥ ŷ → B ∥ A → Φ_B = 0

→ Φ_B =  NBA sin ωt, since Φ_B (t = 0) = 0

Φ_B = − NB₀ πr² sin ωt
ɛ = − dΦ_B / dt = NB₀ πr² ω cos ωt

I =  ɛ / R = (NB₀ωr²/ R) π cos ωt

N  = 15 turns
B₀ = 0.5 Tesla
ω = 300 rad/s
r  = 1 cm =  10⁻²  m
R = 9 ohms

I = (15 ×  0.5 × 300 × 10⁻⁴ / 9) π cos ωt
= (25 × 10⁻³) π sin ωt Ampere
= 25 π sin ωt milliAmpere

Answer: E