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Scilla [17]
3 years ago
7

AM radio signals use amplitude modulation of the radio waves to transmit a signal. A typical wavelength of an AM radio wave is 3

00. meters. What is the frequency of such a radio wave? a) 1.00 µHz b) 1.00 mHz c) 1.00 kHz d) 1.00 MHz e) 1.00 Hz
Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

The frequency of such a radio wave is 1 MHz.

(d) is correct option

Explanation:

Given that,

Wave length = 300 m

We know that,

Speed of light c= 3\times10^{8}\ m/s

We calculate the frequency

Using formula of frequency

c = f\times\lambda

f = \dfrac{c}{\lambda}

Put the value into the formula

f=\dfrac{3\times10^{8}}{300}

f =1\times10^{6}\ Hz

f =1\ MHz

Hence, The frequency of such a radio wave is 1 MHz.

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We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap
marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B=3.7\times 10^{-2} T

Current,I=4.6 A

Diameter of wire,d=0.5 mm=0.5\times 10^{-3} m

Radius of wire,r=\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m

1mm=10^{-3} m

Radius of solenoid,r'=1 cm=1\times 10^{-2} m

1 cm=10^{-2} m

Resistivity of copper,\rho=1.68\times 10^{-8}\Omega m

We know that

R=\frac{\rho l}{A}

Where A=\pi r^2

Using the formula

4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}

l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m

Number of turns of wire=\frac{l}{2\pi r'}

Number of turns of wire=\frac{50.26}{2\pi(1\times 10^{-2}}=800

Hence, the number of turns of the  solenoid,N=799

Magnetic field in solenoid,B=\mu_0 nI

3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6

n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}

n=6404 turns/m

n=\frac{N}{L}

L=\frac{N}{n}

L=\frac{799}{6404}

L=0.125 m=0.125\times 100=12.5 cm

Length of solenoid=12.5 cm

1m=100 cm

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