
An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true.
a) The electron will be accelerated along the axis
b) The electron path will be circular about the axis
c) The electron will experience a force at ${{45}^{\circ }}$to the axis and hence will move in a helical path
d) The electron will continue to move along the axis of the solenoid with uniform velocity.
Answer
570.3k+ views
Hint: The magnetic field lines when current is passed through, become parallel to the axis of the solenoid. It is given in the question that an electron is projected along the axis of the solenoid. Hence we will first determine the force due to the magnetic field produced by the solenoid along the axis and accordingly conclude to an answer from the options provided.
Formula used:
$F=qvBSin\theta $
Complete step by step answer:
Let us say charge ‘q’ is projected into a uniform magnetic field ‘B’ with velocity ‘v’ at an angle $\theta $ . Hence the we can write the expression for the force on the charge due to magnetic field as,
$F=q(v\times B)$.
Since the force is given by the cross product between the vectors, we can write the above expression as,
$F=qvBSin\theta ...(1)$ .
In the above question it is given that the charge is projected along the axis of the solenoid. The magnetic field let us say is also parallel to the direction of the velocity vector of the charge as the field lines are parallel to the axis of the solenoid. Therefore we can imply $\theta ={{0}^{\circ }}$ . But since $sin{{0}^{\circ }}=0$ equation 1 we can write as,
$\begin{align}
& F=qvBSin\theta \\
& \Rightarrow F=qvB(0) \\
& \Rightarrow F=0 \\
\end{align}$
The force due to the magnetic field on the charge will be zero. Hence the charge q will move will neither be accelerated nor will deviate from its path.
So, the correct answer is “Option D”.
Note:
It is to be noted that when the charge enters the magnetic field perpendicular to the direction of motion the path traced by the charge will be circular. Similarly, when the charge enters the magnetic field at angle the charge will trace a helical path.
Formula used:
$F=qvBSin\theta $
Complete step by step answer:
Let us say charge ‘q’ is projected into a uniform magnetic field ‘B’ with velocity ‘v’ at an angle $\theta $ . Hence the we can write the expression for the force on the charge due to magnetic field as,
$F=q(v\times B)$.
Since the force is given by the cross product between the vectors, we can write the above expression as,
$F=qvBSin\theta ...(1)$ .
In the above question it is given that the charge is projected along the axis of the solenoid. The magnetic field let us say is also parallel to the direction of the velocity vector of the charge as the field lines are parallel to the axis of the solenoid. Therefore we can imply $\theta ={{0}^{\circ }}$ . But since $sin{{0}^{\circ }}=0$ equation 1 we can write as,
$\begin{align}
& F=qvBSin\theta \\
& \Rightarrow F=qvB(0) \\
& \Rightarrow F=0 \\
\end{align}$
The force due to the magnetic field on the charge will be zero. Hence the charge q will move will neither be accelerated nor will deviate from its path.
So, the correct answer is “Option D”.
Note:
It is to be noted that when the charge enters the magnetic field perpendicular to the direction of motion the path traced by the charge will be circular. Similarly, when the charge enters the magnetic field at angle the charge will trace a helical path.
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