Question

A charge of 1C is moving in a magnetic field of 0.5Tesla with a velocity of 10m/sec Perpendicular to the field. Force experienced is
A. 5 N
B. 10 N
C. 0.5 N
D. 0 N

Answer 1

Hint:When the charged particle is moving in a magnetic field then the charged particle experiences magnetic force. The magnitude of the magnetic force on the charged particle is determined by Lorentz's force law.

Formula used:
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\]
here \[\overrightarrow F \] is the force acting on the charged particle moving with velocity \[\overrightarrow v \] in a region of magnetic field \[\overrightarrow B \]

Complete step by step solution:
The magnitude of the charge on the moving particle is 1C. The speed of the charged particle is \[10\,m/s\]. The magnetic field strength in the region is 0.5T. It is given that the charged particle is moving perpendicular to the magnetic field, so the angle between the magnetic field and the velocity is \[90^\circ \].

Using Lorentz’s force law, the magnetic force on the moving charged particle is,
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\]
Then the magnitude of the magnetic force on the charged particle is,
\[F = qvB\sin \theta \]
Putting the values, we get
\[F = 1 \times 10 \times 0.5\sin 90^\circ N\]
\[\therefore F = 5N\]
Hence, the moving charge experiences the magnetic force of magnitude of 5N when it is moving in the region containing the magnetic field.

Therefore, the correct option is A.

Note: As we have asked the magnitude of the charged particle then the magnitude of the magnetic field doesn’t depend on the nature of the charged particle. Either of the positive or negative nature of charge with magnitude 1C the particle will experience the same magnitude of force keeping the magnetic field strength same for both.