Question

A cyclotron has an oscillator frequency 12MHz and a dee of radius 50 cm. Calculate the magnetic induction needed to accelerate deuterons of mass $3.3\times {10}^{-27}kg$ and charge $1.6\times {10}^{-19}C$

• A. ${1.55wb/m}^2$
• B. ${2.55wb/m}^2$
• C. ${0.55wb/m}^2$
• D. ${3.55wb/m}^2$

Answer 1

The correct option is A ${1.55wb/m}^2$

Given:

Charge of deutrons ,q= 1.6 ×${10}^{-19}C$

Mass of deutrons, m= 3.3+${10}^{-27}Kg$

Oscillator frequency, n =12 MHz = 12 ×10${}^{6}Hz$

Radius of dee, r = 50cm = 0.5 m

Solution :

$n=\frac{B\times q}{2\times \frac{22}{7}\times m}$

$\Rightarrow B=\frac{n\times 2\times 22\times m}{7\times q}$

$\Rightarrow B=\frac{12\times {10}^6\times 2\times 22\times 3.3\times {10}^{-27}}{7\times 1.6\times {10}^{-19}}$

Therefore, B$=1.55Wb/m^2$

Hence, the correct option is A.