A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R = 50 cm . What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron ?

Question

A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R = 50 cm . What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron ? (A) 0.72 T (B) 0.65 T (C) 0.39 T (D) 0.12 T

Tardigrade Admin · Accepted Answer

Given, f= 12 MHz = 12× 10-6 Hz R = 50 cm= 520 × 10-2 m B = (2π mf/q) = ((12× 106) × 2 × π × (1.67 × 10-27)/(1.6 × 10-19)) = 0.788 ≈ 0.72 T

Q. A cyclotron is operated at an oscillator frequency of $12 M Hz$ and has a dee radius $R = 50 c m$ . What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron ?

0.72 T

0.65 T

0.39 T

0.12 T

Given, $f = 12 M Hz = 12 \times 1 0^{- 6} Hz$
$R = 50 c m = 520 \times 1 0^{- 2} m$
$B = \frac{2 πm f}{q}$
$= \frac{( 12 \times 1 0 ^{6} ) \times 2 \times π \times ( 1.67 \times 1 0 ^{- 27} )}{( 1.6 \times 1 0 ^{- 19} )}$
$= 0.788 \approx 0.72 T$

Q. A cyclotron is operated at an oscillator frequency of 12MHz and has a dee radius R=50cm . What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron ?