A 2. 5 kg copper rod rests on two horizontal rails (see figure below) 2. 2 m apa

Question

A 2. 5 kg copper rod rests on two horizontal rails (see figure below) 2. 2 m apart and carries a current of 50 A from one rail to the other. The coefficient of static friction between rod and rails is 0. 56. What is the smallest magnetic field (not necessarily vertical) that would cause the rod to slide? (Based on the bottom picture, define to the right as the + x - direction and up as the + y - direction. Assume the current in the bottom picture is into the page and the magnetic force will cause the rod will slide to the right. ) magnitude T direction degree counterclockwise from the + x - axis

Explanation / Answer

A 1.0 kg copper rod rests on two horizontal rails 1.0m apart and carries a current of 50 A from one rail to the other. The coeffeicient of static griction between rod and rails is 0.60. What is the smallest magnetic field that would cause the rod to slide?

• ANS

A 1.0 kg copper rod rests on two horizontal rails 1.0m apart and carries a current of 50 A from one rail to the other. The coeffeicient of static griction between rod and rails is 0.60. What is the smallest magnetic field that would cause the rod to slide?

• ANS

A 1.0 kg copper rod rests on two horizontal rails 1.0m apart and carries a current of 50 A from one rail to the other. The coeffeicient of static griction between rod and rails is 0.60. What is the smallest magnetic field that would cause the rod to slide? F = IL x B, where I is current, L is length, B is mag. field. (See the refs.)
IL is across the track, F is along the track, and B is vertical; thus we can say F = ILB.
Assume the rod is 1 kg in mass, then friction force = 0.6mg = 5.88 N, and B = F/(IL) = 5.88/50 = 0.1176 T.
Since you didn't actually provide the mass we have to say B = 0.1176*m T.
The formula F = ILB is based on the Lorentz force law F = qv x B. This means that qv = IL. Note that the units of qv (charge*(L/T)) are the same as IL ((charge/T)*L). So the (large) mobile charge q in that length of rod times its (slow) velocity v (the electron drift rate) = IL. F = IL x B, where I is current, L is length, B is mag. field. (See the refs.)
IL is across the track, F is along the track, and B is vertical; thus we can say F = ILB.
Assume the rod is 1 kg in mass, then friction force = 0.6mg = 5.88 N, and B = F/(IL) = 5.88/50 = 0.1176 T.
Since you didn't actually provide the mass we have to say B = 0.1176*m T.
The formula F = ILB is based on the Lorentz force law F = qv x B. This means that qv = IL. Note that the units of qv (charge*(L/T)) are the same as IL ((charge/T)*L). So the (large) mobile charge q in that length of rod times its (slow) velocity v (the electron drift rate) = IL. F = IL x B, where I is current, L is length, B is mag. field. (See the refs.)
IL is across the track, F is along the track, and B is vertical; thus we can say F = ILB.
Assume the rod is 1 kg in mass, then friction force = 0.6mg = 5.88 N, and B = F/(IL) = 5.88/50 = 0.1176 T.
Since you didn't actually provide the mass we have to say B = 0.1176*m T.
The formula F = ILB is based on the Lorentz force law F = qv x B. This means that qv = IL. Note that the units of qv (charge*(L/T)) are the same as IL ((charge/T)*L). So the (large) mobile charge q in that length of rod times its (slow) velocity v (the electron drift rate) = IL.

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