Question

A uniform field B is acting from south to north and is of magnitude \[1.5\;Wb/{m^2}\]. If a proton having mass of \[1.7 \times {10^{ - 27}}kg\;\] and charge of \[1.6 \times {10^{ - 19}}C\;\] moves in this field vertically downwards with energy \[5 MeV\], then find the force acting on it.
A. \[7.4 \times {10^{12}}N\;\]
B. \[7.4 \times {10^{ - 12}}N\;\]
C. \[7.4 \times {10^{19}}N\;\]
D. \[7.4 \times {10^{ - 19}}N\;\]

Answer 1

Hint: In the given question, we need to find the force acting on protons. For this, we need to use the formula for force experienced by a charged particle in an external magnetic field to get the desired result.




Formula used:
The following formula is used for solving the given question.
The force acting on a particle is given by
 \[F = qB\sqrt {\dfrac{{2K}}{m}} \]
Here, \[F\] is the force, \[q\] is the charge, \[B\] is the magnetic field strength, \[K\] is the kinetic energy and \[m\]is the mass.







Complete answer:
We know that the magnetic force on proton is given by
\[F = qVB\]
Here, \[F\] is the force, \[q\] is the charge, \[B\] is the magnetic field and \[v\] is the velocity.
Also, the kinetic energy is given by
\[K = \dfrac{1}{2}m{v^2}\]
Here, \[K\] is the kinetic energy, \[v\] is the velocity, and \[m\] is the mass.
Thus, we get
\[F = qB\sqrt {\dfrac{{2K}}{m}} \]
\[F = 1.6 \times {10^{ - 19}} \times 1.5 \times \sqrt {\dfrac{{2 \times 5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{1.7 \times {{10}^{ - 27}}kg\;}}} \]
By simplifying, we get
\[F = 7.4 \times {10^{ - 12}}N\;\]
Hence, the force acting on a proton is \[7.4 \times {10^{ - 12}}N\].

Therefore, the correct option is (B).




Note: If the proton and electron are static in the magnetic field or if their velocity, v, and magnetic field, B, are parallel, they will feel zero force. Many students make mistakes in calculation as well as writing the formula of force acting on a proton using kinetic energy. This is the only way through which we can solve the example in the simplest way. Also, it is essential to do calculations carefully to get the correct value of the force.