Question

A particle of charge q and mass m moving with a velocity v along the x-axis enters the region x > 0 with uniform magnetic field B along the \[\hat k\] direction. The particle will penetrate in this region in the x-direction up to a distance d equal to
A. Zero
B. \[\dfrac{{mv}}{{Bq}}\]
C. \[\dfrac{{2mv}}{{Bq}}\]
D. infinity

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. This force is perpendicular to the velocity of the particle, which gives the circular path of the motion.

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.
\[{F_C} = \dfrac{{m{v^2}}}{R}\]
Here \[{F_C}\] is the centrifugal force acting on the particle of mass m moving with speed v in a circular path of radius R.

Complete step by step solution:
The charge on the particle is q. The mass of the particle is m. Using the Cartesian coordinate to represent the velocity vector and the magnetic field vector.
The velocity of the particle is \[\vec v = v\hat i\]
The magnetic field vector is \[\vec B = B\hat k\]

Then using the magnetic force formula the magnetic force vector on the charged particle due to the magnetic field is,
\[\vec F = q\left( {v\hat i \times B\hat k} \right)\]
\[\Rightarrow \vec F = - qvB\hat j\]
So, the magnetic force is having magnitude of qvB and it is acting along the negative y-axis.

On balancing the magnetic force,
\[\dfrac{{m{v^2}}}{r} = qvB\]
\[\Rightarrow r = \dfrac{{mv}}{{Bq}}\]
Then the maximum distance d along the x-axis is the radius of the circular path.
\[d = r\]
\[\therefore d = \dfrac{{mv}}{{Bq}}\]

Therefore, the correct option is B.

Note: We should be careful about the direction in the vector product. We should use the right hand triad in the vector product and determine the force. The direction of the deflection will be in the direction of the magnetic force.