a particle of mass m carrying a charge q is accelerated by a potential difference v it enters perpendicularly - Physics - Moving Charges And Magnetism

Question

a particle of mass m carrying a charge q is accelerated by a
potential difference v it enters perpendicularly in a region of
uniform magnetic field B and executes circular arc of radius R,
then q/m equals?
1) 2v/B2R2
2)v/2BR
pl show the method [the ans- (1)]

Govind · Accepted Answer

Let v be the velocity of the particle of charge q, when accelerated
through potential difference V Then
12mv2=qV      ,
1As the particle describes a semicircular path in the magnetic field
 SoqvB=mv2R⇒v=qBRm     
2Puting this value in equation 1, we get12mqBRm2=qV⇒qm=2VB2R2

a particle of mass m carrying a charge q is accelerated by a potential difference v. it enters perpendicularly in a region of uniform magnetic field B and executes circular arc of radius R, then q/m equals? 1) 2v/B2R2 2)v/2BR pl show the method.[the ans- (1)]

a particle of mass m carrying a charge q is accelerated by a potential difference v. it enters perpendicularly in a region of uniform magnetic field B and executes circular arc of radius R, then q/m equals?
1) 2v/B²R²
2)v/2BR
pl show the method.[the ans- (1)]