Please refer to the uploaded image for the doubt

Asked by arjunsah797 | 16th Mar, 2022, 02:07: PM

Expert Answer:

Fig.(a) shows the setup of experiemental system. A thin aluminium bar is suspended between poles of magnets
so that length of the bar is normal to the direction of magnetic field.This thin aluminium bar is suspended using
a string and two frictionless pulleys as shown in figure.  Some mass pieces  placed on a pan that hangs other side balances
the aluminum bar. Let us use pulleys of negligible mass and negligible friction. Vertical position of the pan is noted using the
moving marker on a graduated scale as shown in figure.

Aluminium bar is connected to a series circuit that consists of battery , ammemeter and rheostat as shown in figure
so that current can be passed through aluminium bar. Magnitude of this current is controlled by Rheostat and measured by ammeter.
 
When current is not passed through aluminium bar , let mi be the mass in the pan .
When current is passed through aluminium bar , since the direction of current is perpendicular to the direction of magnetic field,
magnetic force ( begin mathsize 14px style i l with rightwards arrow on top space cross times B with rightwards arrow on top end style ) acting on the aluminium bar and pushes upward.  Hence the pan moves downwards .
Let us remove mass pieces so that vertical position of the pan becomes same as the position when current is not passed
through aluminium bar. Let mf be the mass in the pan when current is passed through aluminium bat .
 We can calculate the magnetic force from the mass difference ( mi - mf ) as explained below
 
Fig.(b) shows the free body diagram of aluminium bar and pan, when current is not passed through bar.
Forces acting on aluminium bar and pan are shown.
 
At equilibrium , for the bar ,  Ti = m g ................(1)
 
where Ti is the tension in the string when current is not passed through aluminium bar  and ( mg ) is weight of bar.
 
At equilibrium , for the pan ,  Ti = mi g ................(2)
 
where mi is the initial mass in the pan , when current is not passed through aluminium bar.
 
hence we get from eqn.(1) and eqn.(2) ,  m g = mi g  ..........................(3)
 
Fig.(c) shows the free body diagram of aluminium bar and pan, when current is passed through bar.
 
At equilibrium , for the bar ,  Tf = ( m g ) - F ................(4)
 
where Tf is the tension in the string when current is  passed through aluminium bar  and F is magnetic force.
 
At equilibrium , for the pan ,  Tf = mf g ................(5)
 
where mf is the mass in the pan when current is passed through aluminium bar.

hence we get from eqn.(4) and eqn.(5) ,  ( m g ) - F = mf g  ..........................(6)
 
From eqn.(3) and eqn.(6) , we get ,  F = ( mi - mf ) g  ....................... (7)
 
Magnetic force F is given as
 
F = i l B  .....................(8)
 
where i is current passing through aluminium bar , l is length of bar in the magnetic field and B is the magnetic field flux density.
 
We need to notice that l is not the full length of aluminum bar , but part of length in magnetic field . We can take length of magnet as l.
 
From eqn.(7) and (8) , we get
 
B = [ ( mi - mf ) g ] / ( i l ) 

Answered by Thiyagarajan K | 16th Mar, 2022, 11:08: PM