# Proton with kinetic energy

Question:

Proton with kinetic energy of $1 \mathrm{MeV}$ moves from south to north. It gets an acceleration of $10^{12} \mathrm{~m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{-27} \mathrm{~kg}$ )

1. (1) $0.71 \mathrm{mT}$

2. (2) $7.1 \mathrm{mT}$

3. (3) $0.071 \mathrm{mT}$

4. (4) $71 \mathrm{mT}$

Correct Option: 1

Solution:

As we know, magnetic force $F=q v B=m a$

$\therefore \vec{a}=\left(\frac{q v B}{m}\right)$ perpendicular to velocity.

$\therefore$ Also $v=\sqrt{\frac{2 K E}{m}}=\sqrt{\frac{2 \times e \times 10^{6}}{m}}$

$\therefore a=\frac{q v B}{m}=\frac{e B}{m} \sqrt{\frac{2 \times e \times 10^{6}}{m}}$

$\therefore B \simeq \frac{1}{\sqrt{2}} \times 10^{-3} T=0.71 \mathrm{mT}($ approx $)$

$\therefore 10^{12}=\left(\frac{1.6 \times 10^{-19}}{1.67 \times 10^{-27}}\right)^{\frac{3}{2}} \cdot \sqrt{2} \times 10^{3} B$