Question

A charge particle after being accelerated through a potential difference V
enters in a uniform magnetic field and moves in a circle of radius r. If V is
doubled, the radius of the circle will become
(a) 2r
(b) root2r
(c) 4r
(d) r/root2​

Answer 1

Answer:

Force acting on charged particle moving in a magnetic fields is given by:

F=qvB This is equal to the centripetal force acting on the body.

F=Rmv2

Hence R=qBmv

Given initial Radius R=qBmv

Now energy is doubled.

i.e. 21mv22=2∗21mv2

i.e. v2=2v

Hence Radius R2=qB2mv=2R .

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Answer 2

Answer:

· A charge particle after being accelerated through a potential difference V. enters in a uniform magnetic field and moves in a circle of radius r. ... doubled, the radius of the circle will become. (a) 2r.