Question

An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 X ${10}^{-19}$ C, $m_e$ = 9.1 X ${10}^{-31}$ kg)

• A. 2.58 X ${10}^{-4}$m
• B. 3.58 X ${10}^{-4}$m
• C. 2.58 X ${10}^{-3}$m
• D. 3.58 X ${10}^{-3}$m

Answer 1

Answer: (B)

3.58 X ${10}^{-4}$m

$E=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{2E}{m}}$
$r=\frac{mv}{Be}=\frac{m}{Be}\sqrt{\frac{2E}{m}}=\frac{\sqrt{2mE}}{Be}$
$r=\frac{\sqrt{2\times 1800\times 1.6\times {10}^{-19}\times 9.1\times {10}^{-31}}}{1.6\times {10}^{-19}\times 0.4}=3.58\times {10}^{-4}m$