1.

A charge q is accelerated through a potential difference V . It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. Then potential difference required to move it in a circle of radius 2 r is

Answer»

2V
4V
1V
3V

Solution :RADIUS of CIRCULAR path : r = `sqrt(2mqV)/(qB)`
`rpropsqrtV` where B is constant
or `Vpropr^(2)`
`V_(2)/V_(1)=(r_(2)/r_(1))^(2)`
`V_(2)/V=((2r)/r)^(2)4rArrV_(2)=4V`


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