# A copper rod of mass #m# rests on two horizontal rails distance #L# apart and carries a current of #i# from one rail to the other. The coefficient of static friction between rod and rails is #mu_s#. What are the (a) magnitude and (b) angle(relative.....?

##
A copper rod of mass #m# rests on two horizontal rails distance #L# apart and carries a current of #i# from one rail to the other. The coefficient of static friction between rod and rails is #mu_s# . What are the (a) magnitude and (b) angle(relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding?

A copper rod of mass

##### 1 Answer

We are to find the smallest magnetic field that puts the rod of length

The magnetic force generated by current carrying conductor can be so oriented that (1) it must just push the rod horizontally to overcome the force of friction given by

Let the copper rod be oriented along

Let the uniform magnetic field be applied so that it makes an angle

It has two components

#B_darr=Bcostheta# , into the paper ......(1)

#B_x=Bsintheta# .......(2)

Force due the magnetic field will have two components as given by Fleming's right-hand rule,

#F_x=iLB_darr=iLBcostheta# ......(3)

#F_uarr=iLB_x=iLBsintheta# ......(4)

We see that normal reaction becomes

#N=mg-iLBsintheta# ......(5)

When the rod just starts sliding, maximum force of friction, which always opposes the motion, must be equal to component of magnetic force in the

#:.mu_s(mg-iLBsintheta)=iLBcostheta# .....(6)

Solving for magnetic field we get

#B=(mu_smg)/(iL(mu_ssintheta+costheta))# ......(7)

To find out minimum magnetic field we set first differential of this expression with

#(dB)/(d theta)=(mu_smg(mu_scostheta-sintheta))/(iL(mu_ssintheta+costheta)^2)#

#=0# ,

We get

#theta=tan^-1 mu_s# ......(8)

To calculate