Question

The magnetic force acting on a charged particle of $ - 2\mu C$ in a magnetic field of 2T acting in y-direction, when the particle velocity is $\left( {2\widehat i + 3\widehat j} \right) \times {10^6}m{s^{ - 1}}$ is:
A. 8 N in – Z-direction
B. 4 N in Z-direction
C. 8 N in y-direction
D. 4 N in Y-direction

Answer 1

Hint: In these types of questions remember to use the concept of Lorentz force according to which force on a moving charge in magnetic field is given as; $q\left( {\overrightarrow v \times \overrightarrow B } \right)$, using this information will help you to approach the solution of the question.

Complete answer:
According to the given information, it is given that a charged particle of charge $ - 2\mu C$ moving with a velocity of $\left( {2\widehat i + 3\widehat j} \right) \times {10^6}m{s^{ - 1}}$in a magnetic field of 2T which is acting in y direction
So, the given values are \[\overrightarrow V = \left( {2\widehat i + 3\widehat j} \right) \times {10^6}m{s^{ - 1}}\], \[\overrightarrow B = 2\widehat j\]and q = $ - 2\mu C$or $ - 2 \times {10^{ - 6}}C$
We know that by the concept of Lorenz force which says when a charge moves in a magnetic field with some velocity have some charge q then the force acting on that body is given as $\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$
Now substituting the given values in the above formula, we get
$\overrightarrow F = - 2 \times {10^{ - 6}}\left( {\left( {2\widehat i + 3\widehat j \times {{10}^6}} \right) \times 2\widehat j} \right)$
$ \Rightarrow $$\overrightarrow F = - 2 \times \left( {\left( {2\widehat i + 3\widehat j} \right) \times 2\widehat j} \right)$
We know that \[\widehat a \times \left( {\widehat b + \widehat c} \right) = \widehat a \times \widehat b + \widehat a \times \widehat c\]
Therefore, $\overrightarrow F = - 2 \times \left( {2\widehat j \times 2\widehat i + 2\widehat j \times 3\widehat j} \right)$
$ \Rightarrow $$\overrightarrow F = - 2 \times \left( {4\left( {\widehat i \times \widehat j} \right) + 6\left( {\widehat j \times \widehat j} \right)} \right)$
Also, we know that by the law of cross product $\widehat i \times \widehat j = \widehat k$and $\widehat j \times \widehat j = 0$
Therefore, $\overrightarrow F = - 2 \times \left( {4\widehat k + 6\left( 0 \right)} \right)$
$ \Rightarrow $$\overrightarrow F = - 8\widehat k$
Since, we know that $\widehat k$vector represents direction in Z-axis
Therefore, the magnitude of the force on a charge particle will be 8N in – Z direction

So, the correct answer is “Option A”.

Note:
In the above solution we used the term “magnetic force” which can be explained as the force experienced by the electric charge, electric current and magnetic objects due to the magnetic field this force acts perpendicular to the direction of velocity of charge, current or magnetic material as in the above solution the moving charge travels experienced the magnetic force in negative Z-direction due to the magnetic field in which it was moving thus we can say that magnetic force is an vector quantity.