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Afina-wow [57]
3 years ago
12

What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a

magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Physics
1 answer:
Elena L [17]3 years ago
6 0

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

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Answer:

4.5\times 10^{-5} T

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We are given that

Current in wire=40 A

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According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

1 m=100 cm

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B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

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