Question

A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for:

a. its speed
b. its mass.

Answer 1

Answer:

Given that K.E is

1/2mv²

So to find speed v,

Make it subject

K.E= 1.2mv²

However radial force = magnetic force

So mv²/r= qvB

So v subject

V= 2K.E/ qBr that is speed

To find mass

K.E = 1/2mv²

Puy value of v

So KE= 1/2m(2K.E/qBr)

m= (qBr)/2K.E

That is mass

Answer 2

Answer:

m = qbr/v

v = 2k/qbr

Explanation:

When a charged particle enters a magnetic field, it experiences a force that is always perpendicular to the velocity. This force provides a centripetal force, and thus, we have

qvb = mv²/r

if we make m the subject of the formula, we will have

m = qbr/v

Recall that the kinetic energy, KE = ½mv²

Now, let's make v² the subject of formula, we have

v² = 2K/m

now, we substitute for m from the equation we got earlier

v² = 2K / (qbr/v)

v² = 2Kv / qbr, if we simplify further, we have

v = 2k / qbr

Therefore, we can say that the expression for the mass and speed is respectively,

m = qbr/v

v = 2k/qbr