Question

An electron moving with kinetic energy $6.6\times {10}^{-14}J$ enters in a magnetic field of $4\times {10}^{-3}T$ at right angles to it. The radius of its circular path will be nearest to

• A. $100cm$
• B. $75cm$
• C. $25cm$
• D. $50cm$

Answer 1

The correct option is D $50cm$

When a charged particle enters a magnetic field, its path becomes circular whose radius can be found out from the relation.

$\frac{m{v}^2}{r}=Bqv\Rightarrow r=\frac{mv}{Bq}$

$E=\frac{1}{2}m{v}^2=6.6\times {10}^{-14}$

$\frac{1}{2}\frac{m^2{v}^2}{m}=6.6\times {10}^{-14}$

$\Rightarrow m^2{v}^2=2m\times 6.6\times 10-14$

$mv=\sqrt{2m\times 6.6\times {10}^{-14}}$

$=\sqrt{2\times 9.1\times {10}^{-31}\times 6.6\times {10}^{-14}}$

$r=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times 6.6\times {10}^{-14}}}{4\times {10}^{-3}\times 1.6\times {10}^{-19}}$

$=50$cm (approx)