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 = [tex]\frac{1}{2}mv^2[/tex] = 2.5 × 10⁻¹³ J
where, v is the velocity of the electron
or
[tex]\frac{1}{2}\times1.67\times10^{-27}\times v^2[/tex] = 2.5 × 10⁻¹³ J
or
v² = 2.99 × 10¹⁴
or
v = 1.73 × 10⁷ m/s
also,
centripetal force = magnetic force
or
[tex]\frac{mv^2}{r}[/tex] = qvB
q is the charge of the electron
r is the radius of the dipole magnets
on substituting the respective values, we get
[tex]\frac{1.67\times10^{-27}\times1.73\times10^7}{r}[/tex] = 1.6 × 10⁻¹⁹ × 0.75
or
r = 0.2408 m ≈ 24 cm
Hence, the correct answer is 24 cm
The radius of the dipole magnets in the cyclotron is mathematically given as
r=24 cm
Question Parameter(s):
A cyclotron is designed to accelerate protons (mass 1.67 x 10^-27 kg)
Up to kinetic energy of 2.5 x 10^-13 J.
Generally, the equation for the Kinetic energy is mathematically given as
K.E=0.5mv^2
Therefore
0.5*1.67*10^{-27}*v^2 = 2.5 × 10^{-13} J
v = 1.73 × 10⁷ m/s
In conclusion
mv^2/r = qvB
(1.67*10^{-27}*1.73*10^7)/r = 1.6 × 10^{-19}* 0.75
r=24 cm
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