A 200-turn flat coil of wire 30.0 cm in diameter acts as an antenna for FM radio at a frequency of 100 MHz. The magnetic field of the incoming electromagnetic wave is perpendicular to the coil and has a maximum strength of 1.00×10−12 T.

(a) What power is incident on the coil?
(b) What average emf is induced in the coil over one-fourth of a cycle?
(c) If the radio receiver has an inductance of 2.50 μH, what capacitance must it have to resonate at 100 MHz?

a)P = 8.43 × 10⁻¹² W

b) e =5.65 × 10⁻³ V

c) c = 1.01 × 10⁻¹² F

Explanation:

(a) What power is incident on the coil?

diameter(d) = 30.0 cm = 30 × 10⁻² m

Magnetic strength(B₀) = 1 × 10⁻¹² T

Speed of light(c) = 3 × 10⁸ m/s

Permeability of vacuum(u₀) = 4π × 10⁻⁷ N/A² = 12.57 × 10⁻⁷ N/A²

The magnetic field intensity(I) is given as:

The power incident on the coil(P) is given as:

Where A is the coil area = πd²/4 = 3.14d²/4

Therefore,

substituting values,

P = 8.43 × 10⁻¹² W

(b) What average emf is induced in the coil over one-fourth of a cycle

Frequency (f) = 100 MHz = 100 × 10⁶ Hz

emf(e) = ?

Number of turns(N) = 200 turns

Time(T) = 1/f = 1/(100 × 10⁶) = 10⁻⁸ s

dt =

From Faraday's law of electromagnetism:

e =5.65 × 10⁻³ V

c) If the radio receiver has an inductance of 2.50 μH, what capacitance must it have to resonate at 100 MHz?

Capacitance(c) = ?

Inductance(L) = 2.5 μH = 2.5 × 10⁻⁶ H

resonant frequency(f₀) = 100 MHz = 100 × 10⁶ Hz

At resonant frequency, the capacitance impedance and inductive impedance are equal. Therefore:

c = 1.01 × 10⁻¹² F

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explanation: