Question

A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. Because of an electric field along the +y-axis, the charge particle does not change velocity. What is the sign of this particle? 35) A) positive B) negative C) The particle could be either positive or negative. D) None of the above choices is correct.

Answer 1

Explanation:

A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. Because of an electric field along the +y-axis, the charge particle does not change velocity.

As both electric and magnetic field is acting on the charged particle. The Lorentz force will act on it which is given by :

$F=qE+qvB$

$F=q(E+vB)$

q is the charged particle

v is the velocity of the particle

B is the magnetic field

Let the charge particle is positive. The magnetic force will act on it in -y axis using right hand rule. As the direction of electric force is in +y axis, then the electric field is same as electric force i.e. +y axis. As a result both forces cancel out each other.

Similarly, let the charge particle is negative. According to right hand rule, the magnetic force is acting in +y axis. If the charge is negative, the direction of electric field is in opposite direction to the electric force. Thus the electric force is in -y axis. As a result, the both forces cancel out each other.

Hence, the charge can be either positive or negative. Hence, the correct option is (c) "The particle could be either positive or negative".

Answer 2

Because we can take the charge as a common factor in the force equation, we will see that the sign of the charge does not matter. So the correct option is C.

How to find the sign of the particle?

We know that:

• There is a magnetic field on +z.
• There is an electric field in the +y direction.
• The particle moves in the +x direction.

We know that the electromagnetic force is given by:

F = q*E + q*v×B

Where E is the electric field, B is the magnetic field, v is the velocity, and q is the charge.

As you can see, we can take the charge as a common factor, so we get:

F = q*(E + v×B).

Now, if the particle doesn't change its velocity, the force must be zero, and as you can see the force is only zero if the charge is zero, or if the electric field is equal to the cross product between the magnetic field and the velocity.

So the sign of the particle can be either positive or negative, so the correct option is C.

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