
A current carrying long solenoid is placed on the ground with its axis vertical. A proton is falling along the axis of the solenoid with a velocity $v$. When the proton enters into the solenoid, it will
A. Be deflected from its path
B. Be accelerated along the same path
C. Be decelerated along the same path
D. Move along the same path with no change in velocity
Answer
161.4k+ views
Hint: When the charged particle is moving in a magnetic field then it experiences the force due to the magnetic field. The magnetic force is equal to the charge of the particle times the vector product of the velocity and the magnetic field.
Formula used:
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\]
Here \[\overrightarrow F \] is the magnetic force vector acting on the charged particle which is moving with velocity \[\overrightarrow v \] in a magnetic field \[\overrightarrow B \].
Complete step by step solution:
When the current is flowing in the long solenoid then there is a magnetic field produced inside the solenoid. The direction of the magnetic field lines is almost parallel to the axis of the solenoid when the length of the solenoid is long enough. So, the magnetic field direction inside the solenoid is parallel to the axis of the solenoid.
It is a given that a proton is falling along the axis of the solenoid with speed v. So, the direction of the velocity is parallel to the axis of the solenoid. Hence, the angle between the magnetic field due to solenoid and the velocity of the proton is zero or 180°. As we know that the vector product of two parallel or antiparallel vectors is zero. Hence, the magnetic force on the moving proton is zero. When the force acting on the particle is zero then acceleration will be zero, i.e. the velocity will be constant.
Therefore, the correct option is D.
Note: We should be aware of the fact that when a particle is falling then the gravitational force also acts on it which may cause the change in speed of the particle. In this question it was not mentioned to consider the gravitational force, we assume the motion is in gravity free space.
Formula used:
\[\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)\]
Here \[\overrightarrow F \] is the magnetic force vector acting on the charged particle which is moving with velocity \[\overrightarrow v \] in a magnetic field \[\overrightarrow B \].
Complete step by step solution:
When the current is flowing in the long solenoid then there is a magnetic field produced inside the solenoid. The direction of the magnetic field lines is almost parallel to the axis of the solenoid when the length of the solenoid is long enough. So, the magnetic field direction inside the solenoid is parallel to the axis of the solenoid.
It is a given that a proton is falling along the axis of the solenoid with speed v. So, the direction of the velocity is parallel to the axis of the solenoid. Hence, the angle between the magnetic field due to solenoid and the velocity of the proton is zero or 180°. As we know that the vector product of two parallel or antiparallel vectors is zero. Hence, the magnetic force on the moving proton is zero. When the force acting on the particle is zero then acceleration will be zero, i.e. the velocity will be constant.
Therefore, the correct option is D.
Note: We should be aware of the fact that when a particle is falling then the gravitational force also acts on it which may cause the change in speed of the particle. In this question it was not mentioned to consider the gravitational force, we assume the motion is in gravity free space.
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