Question

A particle with \[{10^{ - 11}}\] coulomb of charge and \[{10^{ - 7}}\] kg mass is moving with a velocity of \[{10^8}\] m/s along the y-axis. A uniform static magnetic field \[B = 0.5\] Tesla is acting along the x-direction. The force on the particle is
A. \[5 \times {10^{ - 11}}N\] along \[\widehat i\]
B. \[5 \times {10^3}N\] along \[\widehat k\]
C. \[5 \times {10^{ - 11}}N\] along \[ - \widehat j\]
D. \[5 \times {10^{ - 4}}N\] along \[ - \widehat k\]

Answer 1

Hint: When the charged particle is moving in the magnetic field then due to interaction of the charged particle with the magnetic field the charged particle experiences the force. The magnetic force is proportional to the product of the charge, speed and the magnetic field strength.

Formula used:
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
Here \[\vec F\] is the magnetic force vector, \[\vec v\] is the velocity of the charged particle and \[\vec B\] is the magnetic field in the region.

Complete step by step solution:
The charge on the particle is \[q = {10^{ - 11}}C\]
The mass of the particle is \[m = {10^{ - 7}}kg\]
Using the Cartesian coordinate to represent the velocity vector and the magnetic field vector,
The velocity of the particle is \[\vec v = \left( {{{10}^8}m/s} \right)\hat j\]
The magnetic field vector is \[\vec B = \left( {0.5T} \right)\hat i\]

Then using the magnetic force formula the magnetic force vector on the charged particle due to the magnetic field is,
\[\vec F = {10^{ - 11}}\left( {{{10}^8}\hat j \times 0.5\hat i} \right)N\]
As using the right hand triad we know that \[\hat j \times \hat i = - \hat k\]
\[\vec F = \left( {5 \times {{10}^{ - 4}}N} \right)\left( { - \hat k} \right)\]
So, the magnitude of the magnetic force acting on the charged particle when moving through the magnetic field is \[5 \times {10^{ - 4}}N\] and the direction is along \[ - \hat k\].

Therefore, the correct option is D.

Note: We should be careful about the nature of the charge. While substituting the charge in the magnetic force formula, we need to put the charge, not only the magnitude of the charge on the particle. We need to use the right hand triad throughout the vector product.