Question

A \(10\ eV\) electron is circulating in a plane at right angles to a uniform field at magnetic induction \({10}^{- 4}\ Wb/m^2\) (= \(1.0\) gauss), the orbital radius of electron is

• A. 11 cm
• B. 18 cm
• C. 12 cm
• D. 16 cm

Answer 1

The Correct Option is A

Kinetic energy of electron \(\bigg( \frac{1}{2} \times m v^2 \bigg) = 10 \,eV\) 
and magnetic induction (B) = \({10}^{-4} Wb / m^2\) 
Therefore \(\frac{ 1}{2} (9.1 \times {10}^{-31}) v^2 = 10 \times (1.6 \times {10}^{-19})\) 

or, \(v^2 = \frac{ 2 \times 10 \times (1.6 \times {10}^{-19})}{ 9.1 \times {10}^{-31}} = 3.52 \times {10}^{12}\) 

or , \(v= 1.876 \times {10}^6 \,m\). 

Centripetal force = \(\frac{ m v^2}{r} = Bev .\) 

Therefore \(r= \frac{ mv}{Be} = \frac{ (9.1 \times {10}^{-31}) \times (1.876 \times {10}^6)}{ {10}^{-4} \times (1.6 \times {10}^{-19})}\)
\(= 11 \times {10}^{-2} m\)
\(= 11\, cm\).

So, the correct option is (A): 11 cm