Question

charge partiele after being accelerated through a potential difference V
enters in a uniform magnetic field and moves in a circle of radius r. If Vis
doubled, the radius of the circle will become
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Answer 1

If potential difference V is doubled, the radius of the circle will become 1.414  r

1. The kinetic energy of a accelerated charge is given as
2. 1/2mv^2=qV here v is the velocity ,m is the mass q is the charge and V is the potential difference
3. the velocity is $v=\sqrt{\frac{2qV}{m} }$
4. The radius of circle formed by a charge q  in a uniform magnetic field is given as
5. $r=\frac{mv}{Bq}$ now putting the value of v in this equation
6. $r=\frac{m}{Bq}\sqrt{\frac{2qV}{m} }$ so r is directly proportional to $\sqrt{V}$
7. Now when the V is doubled the radius become 1.414 r