Grade 11Magnetism
Ques.: A rod of length 'l' rotates with a small but uniform angular velocity 
about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is :
about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is :Answer: 1/8(B l^2) [Reference: H.C.Verma II
B l^2) [Reference: H.C.Verma II Though I've calculated the answer correct by Integration method,
but can anyone please solve this step by step by proceeding from this method -
(mv^2)/r = F(B) - F(E)
=> (mv^2)/r = evB - eE
=> mr^2 = erB - e 2 V / l ............. ????
^2 = erB - e 2 V / l ............. ????Can anyone plzzz...
