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2 years ago

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A conducting bar slides without friction on two parallel horizontal rails that are 50 cm apart and connected by a wire at one en

d. The resistance of the bar and the rails is constant and equal to 0.10 Ω. A uniform magnetic field is perpendicular to the plane of the rails. A 0.080-N force parallel to the rails is required to keep the bar moving at a constant speed of 0.50 m/s. What is the magnitude of the magnetic field?

1 answer:

MatroZZZ [7]2 years ago

7 0

Answer:

0.25 T

Explanation:

F = Force required to keep the bar moving = 0.080 N

B = magnitude of magnetic field = ?

L = length of the bar = 50 cm = 0.50 m

v = speed of the bar = 0.50 m/s

R = resistance of the bar =0.10 Ω

Force is given as

$F = \frac{B^{2}L^{2}v}{R}$

$0.08 = \frac{B^{2}(0.50)^{2}(0.50)}{0.10}$

B = 0.25 T

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Answer:

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The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

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$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

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$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

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