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Paraphin [41]
4 years ago
15

If a single circular loop of wire carries a current of 62 A and produces a magnetic field at its center with a magnitude of 1.20

10-4 T, determine the radius of the loop.
Physics
1 answer:
s2008m [1.1K]4 years ago
3 0

Given Information:  

Current in loop = I = 62 A

Magnitude of magnetic field = B = 1.20x10⁻⁴ T

Required Information:  

Radius of the circular loop = r = ?  

Answer:  

Radius of the circular loop = 0.324 m

Explanation:  

In a circular loop of wire with radius r and carrying a current I  induces a magnetic field B which is given by

B = μ₀I/2r

Please note that for an infinitely straight long wire we use 2πr whereas for circular loop we use 2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space

Re-arranging the equation yields

r = μ₀I/2B

r = 4πx10⁻⁷*62/2*1.20x10⁻⁴

r = 0.324 m

Therefore, the radius of this circular loop is 0.324 m

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The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
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Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

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How does a balanced Chemical Equation Satisfy the Law of Conservation of Mass?
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The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th
jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = 130\times 6.894

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or,

   = 896.22\times 10^3 \ Pa

Z₂ = 10ft

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According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  \frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2

On substituting the values, we get

⇒  \frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2

⇒  \frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05

⇒  P_2=866330 \ P_a

i.e.,

⇒       =866330\times 0.000145

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Answer:

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Explanation:

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