Question

A positron with kinetic energy 3.00 keV is projected into a uniform magnetic field of magnitude 0.140 T, with its velocity vector making an angle of 88.0° with the field. Find (a) the period, (b) the pitch p, and (c) the radius r of its helical path.

ASAP

Answer 1

The velocity vector makes an angle with the field means we have two components of velocity

so,

for motion of charged particle in magnetic field

we know that

r = mv / qB

where v is the horizontal component, v sin

so,

r = mvsin / qB -------------- (1)

so,

time period, T is given as

T = 2r / vsin ------------------- (2)

put (1) in (2), we get

T = 2 m / qB

T = 2 * 9.1e-31 / 1.6e-19 * 0.140

T = 2.55e-10 seconds

----------------------------------------------

(b)

Now ....what is pitch ??

it is the distance traveled along magnetic field line in one time period

so,

d = v * t

here, v is component of velocity along the field lines, v cos

we need to find v,

v = sqrt ( 2 * K.E / m)

v = sqrt ( 2 * 3e3 * 1.6e-19 / 9.1e-31)

v = 3.247e7 m/s

therefore,

d = 3.247e7 * cos 88 * 2.55e-10

d = 2.89e-4 m

-----------------------------------

(c)

we just found 'r' in (a)

r = mvsin / qB

r = 9.1e-31 * 3.247e7 * sin 88 / 1.6e-19 * 0.14

r = 1.318e-3 m