Question

A cyclotron when being used to accelerate positive ions?(mass=6.7 into 10^-27,charge=3.2 into 10^-19 C) has magnetic field of pi/2 T. What must be the value of the frequency of the applied alternating electric field to be used in it?

Answer 1

Answer:

$\boxed{ \huge{ \mathfrak{f_a = 11.94 \ MHz}}}$

Given:

Mass (m) = $\sf 6.7 \times 10^{-27}$ kg

Charge (q) = $\sf 3.2 \times 10^{-19}$ C

Magnetic field (B) = $\sf \frac{\pi}{2}$ T

To Find:

Frequency of the applied alternating electric field ($\sf f_a$)

Explanation:

Formula:

$\boxed{ \bold{\sf f_a = \frac{qB}{2\pi m}}}$

Substituting values of q, B & m in the equation:

$\sf \implies f_a = \frac{3.2 \times {10}^{ - 19} \times \frac{\pi}{2} }{2\pi \times 6.7 \times {10}^{ - 27} }$

$\sf \implies f_a = \frac{3.2 \times {10}^{ - 19} \times \cancel{\pi} }{2 \cancel{\pi} \times 6.7 \times {10}^{ - 27} \times 2 }$

$\sf \implies f_a = \frac{ \cancel{4} \times 0.8 \times {10}^{ - 19} }{\cancel{4} \times 6.7 \times {10}^{ - 27} }$

$\sf \implies f_a = \frac{ 0.8 \times {10}^{ - 19} }{ 6.7 \times {10}^{ - 27} }$

$\sf \implies f_a = \frac{ 0.119 \times {10}^{ - 19} }{ {10}^{ - 27} }$

$\sf \implies f_a = 0.1194 \times {10}^{ - 19 - ( - 27)}$

$\sf \implies f_a = 0.1194 \times {10}^{ - 19 + 27}$

$\sf \implies f_a = 0.1194 \times {10}^{8}$

$\sf \implies f_a = 11.94 \times {10}^{6} \: Hz$

$\sf \implies f_a = 11.94 \: MHz$

$\therefore$

Frequency of the applied alternating electric field ($\sf f_a$) = 11.94 MHz

Answer 2

Answer:

f = qB/2πm

By putting values in equation we get

f = 1.2 × 10^7 Hz