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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 : Answer: 1/8( 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 = er B - e 2 V / l ............. ???? Can anyone plzzz...


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 :
 
Answer: 1/8(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   ............. ????
 
Can anyone plzzz...


Grade:11

1 Answers

ROSHAN MUJEEB
askIITians Faculty 833 Points
3 years ago
et us consider a small elementdxat a distancexfrom the centre of the rod rotating with angular velocityωabout its perpendicular bisector. The emf induced in the rod because of this small element is given by
de=Bvl=Bωxdxde=Bvl=Bωxdx

The emf induced across the centre and end of the rod is given by
∫de=∫l/20Bωxdx⇒E=Bω[x22]l/20⇒E=18Bωl2

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