Deuteron vs Proton in Magnetic Field

Kinetic Energy of Proton in Magnetic Field (1991)

Question: A deuteron (KE = 50 keV) orbits with radius 0.5 m in a magnetic field. What is the KE of a proton with the same radius and field?

(a) 25 keV

(b) 50 keV

(c) 200 keV

(d) 100 keV

Solution

Step 1: Radius Formula
For circular motion in a magnetic field: \[ r = \frac{mv}{qB} \implies v = \frac{qBr}{m} \]

Step 2: Relate Velocities
Deuterium (\( m_d = 2m_p \)) and proton (\( m_p \)) have the same radius \( r \) and charge \( q \): \[ \frac{m_d v_d}{qB} = \frac{m_p v_p}{qB} \implies 2m_p v_d = m_p v_p \implies v_p = 2v_d \]

Step 3: Kinetic Energy Ratio
Kinetic energy \( E = \frac{1}{2}mv^2 \):
For deuteron: \( E_d = \frac{1}{2}(2m_p)v_d^2 = m_p v_d^2 = 50 \, \text{keV} \)
For proton: \( E_p = \frac{1}{2}m_p(2v_d)^2 = 2m_p v_d^2 = 2 \times 50 \, \text{keV} = 100 \, \text{keV} \)

Answer: Option (d) \( 100 \, \text{keV} \) is correct.