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 incomi

Question

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?

in progress 0
RI SƠ 5 months 2021-08-25T00:08:37+00:00 2 Answers 10 views 0

Answers ( )

  1. Jezebel
    0
    2021-08-25T00:09:44+00:00
    Reply

    Answer:

    a) 8.4×10^-12W

    b) 5.65 ×10^ -3V

    c) 1.01 ×10^ -12 F

    Explanation: Given:

    Radius r = d/2 = 30/2=15cm= 0.15m

    Maximum feild strength = Bo= 10^-12 T

    Permiability of vacuum, Uo = 4pi×10^-7

    I = CBo/2Uo = (3 ×10^8×(10^-12)^2)/ (4 ×3.142×10^-7)

    I = 1.19×10^-10W/m^2

    The intensity I = > P/A = p = IA = I(4pi^2)

    P = 1.19×10^-10 × 3.142 × (0.15^2)

    P = 8.4 × 10^-12W

    b) Given: f = 100Mhz= 100×10^6hz

    Number of turns in the coil, N = 200

    Time of cycle, T = 1/f = 1/ 100×10^6

    T = 10^-8 seconds

    Change in t= 1/4T = 1/4×10^-8 = 2.5 ×10^-9sec

    E = (N×change in B×A)/ change in t

    E = (NBopir^2)/change in t

    E = (200 ×10^-12× 3.142 ×(0.15^2))/ (2.5×10^-9)

    E = 5.65 ×10^-3V

    c) Self inductance L = 2.5UH=2.5×10^-6H

    Resonant frequency, fo= 1/(2pisqrt(LC))

    C= 1/(4×3.142×( 2.5×10^-6)× (100×10^6)^2

    C = 1.01×10^-12F

  2. thucuc
    0
    2021-08-25T00:10:10+00:00
    Reply

    Answer:

    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:

    I= \frac{cB_{o}^{2} }{2u_{o} }

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

    P=IA

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

    Therefore, P= \frac{cB_{o}^{2} }{2u_{o} }*\frac{3.14d^{2} }{4}

    substituting values,

    P= \frac{3*10^{8}( 1*10^{-12} )^{2} }{2*12.57*10^{-7} }*\frac{3.14*(30*10^{-2}) ^{2} }{4}

    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 = \frac{1}{4}T=\frac{1}{4} *10^{-8} = 2.5*10^{-9} s

    From Faraday’s law of electromagnetism:

    e=N\frac{dO_{m} }{dt} = N\frac{B_{0}A }{dt} \\e= 200(\frac{10^{-12}*\frac{3.14*(30*10^{-2})^{2} }{4} }{2.5*10^{-9} })

    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:

    f_{0}=\frac{1}{2*3.14\sqrt{LC} }

    c=\frac{1}{(2*3.14)^{2}Lf_{0}^{2} } \\c=\frac{1}{(2*3.14)^{2}*2.5*10^{-6}* (100*10^{6}) ^{2} }

    c = 1.01 × 10⁻¹² F

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )