Question

A singly charged positive ion that has a mass of 6.07 ? 10-27 kg moves clockwise with a speed of 1.18 ? 104 m/s. the positively charged ion moves in a circular path that has a radius of 2.75 cm. find the direction and strength of the uniform magnetic field through which the charge is moving. (hint: the magnetic force exerted on the positive ion is the centripetal force, and the speed given for the positive ion is its tangential speed.)

Answer 1

B = 0.018 T Ans,

Since, it is moving in a circular path, thus, centripetal force will act on it i.e.
F =
`( mv^(2) )/(r)`
where, m is the mass of the object, v is the velocity and r is the radius of circular path.

And, since a positive charge is moving, it will create magnetic force which is equal to F = qvB
where q is the charge, v is the velocity of the particle and B is the magnetic field.
Now, the two forces will be equal,
i.e.
`( mv^(2) )/(r)` = qvB
⇒
`( mv )/(r)` = qB
⇒B =
`( mv )/(qr)`
putting the values, we get,

use q = 1.6 * 10^ -19
⇒ B = 0.018 T