Question

A cyclotron is designed to accelerate protons (mass 1.67 x 10^-27 kg) up to a kinetic energy of 2.5 x 10^-13 J. If the magnetic field in the cyclotron is 0.75T, what is the radius of the dipole magnets in the cyclotron? 14cm 17cm 24cm 5.8cm

Answer 1

Answer:

24 cm

Explanation:

Given:

Mass of proton = 1.67 × 10⁻²⁷ Kg

kinetic energy = 2.5 × 10⁻¹³ J

magnetic field in the cyclotron, B = 0.75 T

Now,

Kinetic energy = $\frac{1}{2}mv^2$  = 2.5 × 10⁻¹³ J

where, v is the velocity of the electron

or

$\frac{1}{2}\times1.67\times10^{-27}\times v^2$  = 2.5 × 10⁻¹³ J

or

v² = 2.99 × 10¹⁴

or

v = 1.73 × 10⁷ m/s

also,

centripetal force = magnetic force

or

$\frac{mv^2}{r}$  = qvB

q is the charge of the electron

r is the radius of the dipole magnets

on substituting the respective values, we get

$\frac{1.67\times10^{-27}\times1.73\times10^7}{r}$  = 1.6 × 10⁻¹⁹ × 0.75

or

r = 0.2408 m ≈ 24 cm

Hence, the correct answer is 24 cm