Thus, final torque tf = m B = 10 ´ 2 = 20 N m (d)  From Newton’s second law,
I  
where I  is the moment of inertia of the coil From chain rule,
d
d
d
d
d
d
d
d


t
t



 


Using this,
I  
d
sin
d
 m B




Integrating from q = 0 to q = p/2,
Example 4 12
(a) A current-carrying circular loop lies on a smooth horizontal plane