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mojhsa [17]
4 years ago
5

A flat, circular loop has 20 turns. The radius of the loop is 10.0 cm and the current through the wire is 0.50 A. Determine the

magnitute of magnetic field at the center of the loop.
Physics
1 answer:
nata0808 [166]4 years ago
8 0

Answer:B=6.28\times 10^{-5} T

Explanation:

Given

No of turns n=20 turns

radius of Loop r=10 cm=0.1 m

current i=0.5 A

Magnetic Field at center of loop is given by

B=\frac{\mu _0nI}{2r}

B=\frac{4\pi \times 10^{-7}\times 20\times 0.5}{2\times 0.1}

B=628.4\times 10^{-7} T

B=6.28\times 10^{-5} T

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What is the accletation of a 1500kg with a net force of 7500 N
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3 years ago
In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3
Archy [21]

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Newton

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4 0
4 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
3 years ago
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