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kobusy [5.1K]
3 years ago
10

IN a physics lab, a student discovers that the magnitude of the magnetic field at a certain distance from a long wire is 4.0μT.

If the wire carries a current of 5.0 A, what is the distance of the magnetic field from the wire?
Physics
1 answer:
Alinara [238K]3 years ago
7 0

Answer:

0.25 m

Explanation:

The intensity of the magnetic field around a current-carrying wire is given by:

B=\frac{\mu_0 I}{2 \pi r}

where

where

\mu_0 = 4\pi \cdot 10^{-7}Tm/A is the permeabilty of free space

I is the current

r is the distance from the wire

In this problem, we know:

B=4 \mu T=4 \cdot 10^{-6} T is the magnetic field

I=5.0 A is the current in the wire

Re-arranging the equation, we can find the distance of the field from the wire:

r=\frac{\mu_0 I}{2 \pi B}=\frac{(4\pi \cdot 10^{-7})(5.0 A)}{2\pi(4\cdot 10^{-6} T)}=0.25 m

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Answer:

2.5 × 10-⁴¹ Ns

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Impulse

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borishaifa [10]

Answer:

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Explanation:

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2 years ago
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

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3 years ago
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Answer:

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Explanation:

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