Ask Question
13 April, 03:38

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?

+2
Answers (1)
  1. 13 April, 03:43
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers