What uniform magnetic field applied perpendicular to a beam of electrons moving at $1.3 \times 10^6 m/s$, is required to make the electrons travel in a circular arc of radius 0.35 m?

$2.1 \times 10^{-5} T$

The correct answer is Option (3) → $2.1 \times 10^{-5} T$

The magnetic force is given by -

$F_{magnetic}=evB$

$F_{centripetal}=\frac{mv^2}{r}$

and,

$F_{magnetic}=F_{centripetal}$

$evB=\frac{mv^2}{r}$

$⇒B=\frac{mv}{er}=\frac{(9.1×10^{-31})(1.3×10^6)}{(1.6×10^{-19})(0.35)}$

$≃2.1 \times 10^{-5} T$