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A 0.150-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.510 m apart. If the coefficient of kinetic friction between the rod and rails is 0.160, what vertical magnetic field is required to keep the rod moving at a constant speed?

Respuesta :

jolis1796

Answer:

B = 0.0307 T = 30.74 mT

Explanation:

Given

m = 0.150 kg

I = 15.0 A

d = 0.510 m

μk = 0.16

B = ?

Balancing the forces on the rod in the j direction

N - m*g = 0   ⇒   N = m*g

and in the i direction

I*d*B - μk*N = 0     ⇒      B = μk*N / (I*d)

⇒      B = μk*m*g / (I*d)

⇒      B = (0.16)*(0.150 kg)*(9.8 m/s²) / (15.0 A*0.510 m)

⇒      B = 0.0307 T = 30.74 mT

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