question_answer

Proton with kinetic energy of 1 MeV 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\[1.6\times {{10}^{27}}kg\]) [JEE MAIN Held On 08-01-2020 Morning]

A) \[0.071\text{ }mT\]

B) \[0.71\text{ }mT\]

C) \[71\text{ }mT\]

D) \[7.1\text{ }mT\]

Correct Answer: B

Solution :

[b]

Magnetic Force \[F=qvB\]

\[\therefore \vec{a}=\left( \frac{qvB}{m} \right)\] Perpendicular to velocity.

\[\therefore \]Also             \[v=\sqrt{\frac{2k}{m}}=\sqrt{\frac{2\times e\times {{10}^{6}}}{m}}\]

\[\therefore a=\frac{qvB}{m}=\frac{eB}{m}\sqrt{\frac{2\times e\times {{10}^{6}}}{m}}\]

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

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