Question

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

• A. $0.71mT$
• B. $7.1mT$
• C. $71mT$
• D. $0.071mT$

Answer 1

The correct option is A

$0.71mT$

Step 1: Given Data:

Kinetic Energy $=1MeV=1\times {10}^{6}eV$

$1eV=1.6\times {10}^{-19}J$

So, kinetic energy will be in terms of joule-

$1\times {10}^{6}eV=1.6\times {10}^{-13}J$

Mass of proton $\left(m_p\right)=1.6\times {10}^{-27}kg$

Charge on proton $q=1.6\times {10}^{-19}C$

Acceleration of particle $={10}^{12}m/s^2$

Step 2: Formula Used:

Kinetic Energy$=\frac{1}{2}m{v}^2$

Lorentz Formula$\left(F\right)=qvB\sin \theta$

Lorentz Formula$\left(F\right)=qvB\sin {90}^{\circ }=qvB$

Where $q$is an electric charge on the particle

$V$ is the velocity of the particle

$B$ is the magnetic field

Force$\left(F\right)=m\times a$

From the above two formulas

$qvB=ma$……………………(1)

Step 3: Calculating the velocity of the proton:

The kinetic energy using the formula,

$\begin{array}{rcl}KE& =& \frac{1}{2}{m}_p{v}^2\\ 1.6\times {10}^{-13}& =& \frac{1}{2}\times 1.6\times {10}^{-27}\times v^2\\ v^2& =& 2\times {10}^{14}\\ v& =& \sqrt{2}\times {10}^{7}m/s\end{array}$

The velocity of the proton is $v=\sqrt{2}\times {10}^{7}m/s$
Step 4: Calculating the value of the magnetic field:

Using the equation (1)-

The magnetic field can be calculated as-

$\begin{array}{rcl}Bqv& =& m_{p}a\\ B& =& \frac{m_{p}a}{qv}\\ B& =& \frac{1.6\times {10}^{-27}\times {10}^{12}}{1.6\times {10}^{-19}\times \sqrt{2}\times {10}^7}\\ & =& 0.71\times {10}^{-3}T\\ & =& 0.71mT\end{array}$

Hence, option (A) is the correct answer.