Question

A 0.66 kg copper rod rests on two horizontal rails 0.66 m apart and carries a current of 65 A from one rail to the other. The coefficient of static friction between rod and rails is 0.54. What is the smallest magnetic field (not necessarily vertical ) that would cause the rod to slide?

a)What is the angle of B from the vertical? (deg)

b) What is the magnitude of B?

Answer 1

F has an eastward component Fx and an upward component Fy, which can be related to the components of the magnetic field once we assume a direction for the current in the rod. Thus, again for definiteness, we assume the current flows northward. Then, by the right-hand rule, a downward component (Bd) of B will produce the eastward Fx, and a westward component (Bw) will produce the upward Fy. Specifically,

Fx=iL(Bd) ,Fy=iL(Bw)

Considering forces along a vertical axis, we find

Fn=mg-Fy=mg-iL(Bw)

So that

f=fs,max= µs(mg-iLBw)

It is on the verge of motion, so we set the horizontal acceleration to zero:

Fx-f=0

I*L*Bd= µs(mg-iLBw)

The angle of the field components is adjustable, and we can minimize with respect to it . Defining the angle by Bw = B* sinӨ and Bd=B*cosӨ (which means Ө measured from a vertical axis) measured from a vertical axis) and writing the above expression in these terms, we obtain

iLBcosӨ= µs(mg-iLBsinӨ) which gives B= µs*mg/iL(cosӨ+ µs*sinӨ)

which we differentiate and set the result to zero.this provived the angle :

Ө=tan^-1(µs)= tan^-1(0.54)=28.36=28.4

Consequently,

Bmin=0.54(0.66kg)(9.8m)/(65)(0.66)(cos28.4+0.54sin28.4)=0.071 T