Question

The magnetic field required to accelerate deuterons in a cyclotron operated at a frequency of 28 MHz is (Mass of proton $=1.67\times10^{-27}kg$)

• A. 7.348 T
• B. 0.917 T
• C. 1.837 T
• D. 3.674 T \bigskip

Hint:

The cyclotron frequency depends only on the charge-to-mass ratio \(\frac{q}{m}\) and the magnetic field \(B\). It does not depend on the radius of the orbit.

Answer 1

The Correct Option is C

Concept: A cyclotron accelerates charged particles using a constant magnetic field and a high-frequency alternating electric field. The cyclotron frequency is given by \[ f=\frac{qB}{2\pi m} \] where \[ f=\text{frequency of revolution} \] \[ q=\text{charge of particle} \] \[ m=\text{mass of particle} \] \[ B=\text{magnetic field} \] For a deuteron, \[ m_d=2m_p \] approximately.

Step 1: Write the cyclotron frequency formula.
\[ f=\frac{qB}{2\pi m} \] Rearranging for magnetic field, \[ B=\frac{2\pi mf}{q} \]

Step 2: Calculate the mass of deuteron.
Given, \[ m_p=1.67\times10^{-27}kg \] Therefore, \[ m_d=2m_p \] \[ m_d=3.34\times10^{-27}kg \]

Step 3: Substitute the numerical values.
Frequency: \[ f=28\times10^{6}Hz \] Charge of deuteron: \[ q=1.6\times10^{-19}C \] Hence, \[ B= \frac{2\pi(3.34\times10^{-27})(28\times10^6)} {1.6\times10^{-19}} \]

Step 4: Perform the calculations carefully.
\[ 2\pi\times3.34\times28 \approx587.6 \] and \[ 10^{-27+6+19}=10^{-2} \] Thus, \[ B\approx \frac{587.6\times10^{-2}}{1.6} \] \[ B\approx1.837T \]

Step 5: State the required magnetic field.
Therefore, \[ \boxed{B=1.837T} \] Hence the correct option is \[ \boxed{\text{(C)}} \]