Question

A 1.20 kg copper rod resting on two horizontal rails 0.90 m apart carries a

current I = 55.0 A from one rail to the other. The coefficient of static friction
between the rod and rails is μs= 0.60.
(a) What is the smallest vertical magnetic field B that would cause the rod to
slide?

(b) Suppose a B field is directed at some angle to the vertical φ, with the current
along the rod directed into the page, as shown. Find an expression for B as a
function of φ for the case when the rod is just on the verge of beginning to slide.

(c) Find the value of φ which yields the smallest value of B that would cause
the rod to slide, together with the corresponding value of B.

Answer 1

Step-by-step explanation:

(a) In order for the copper rod to slide, the magnetic force on it must be greater than the maximum static friction force. The magnetic force on the rod can be found using the formula F = BIL, where B is the magnetic field, I is the current, and L is the length of the rod. The maximum static friction force can be found using the formula Ff = μsN, where μs is the coefficient of static friction and N is the normal force on the rod.

Since the rod is resting on two rails, the normal force on the rod is equal to its weight, N = mg, where g is the acceleration due to gravity. Therefore, the condition for the rod to slide is:

BIL > μs mg

Solving for B, we get:

B > μs mg / IL

Substituting the given values, we get:

B > (0.60)(1.20 kg)(9.81 m/s^2) / (0.90 m)(55.0 A)

B > 0.077 T

Therefore, the smallest vertical magnetic field that would cause the rod to slide is 0.077 T.

(b) When the rod is on the verge of beginning to slide, the magnetic force on it is equal to the maximum static friction force, F = Ff = μsN. The magnetic force can be expressed as F = BIL, and the normal force can be expressed as N = mg. Therefore, we have:

BIL = μs mg

Solving for B, we get:

B = μs mg / IL

But we also know that the angle between the magnetic field and the vertical is given by φ, so we can express L in terms of φ using the formula L = d/sinφ, where d is the distance between the rails. Therefore, we have:

B = μs mg sinφ / Id

(c) To find the value of φ that yields the smallest value of B, we need to minimize the expression for B with respect to φ. Taking the derivative of B with respect to φ, we get:

dB/dφ = μs mg cosφ / Id sin^2φ

Setting this derivative equal to zero and solving for φ, we get:

tanφ = μs mg / Id

Substituting the given values, we get:

tanφ = (0.60)(1.20 kg)(9.81 m/s^2) / (0.90 m)(3000000 kg)(9.00 x 10^-7 m)

tanφ ≈ 0.12

Taking the arctan of both sides, we get:

φ ≈ 6.87°

Substituting this value of φ back into the expression for B, we get:

B = μs mg sinφ / Id

B ≈ 0.030 T

Therefore, the smallest value of B that would cause the rod to slide is approximately 0.030 T, when the magnetic field is at an angle of 6.87° to the vertical.