0

Consider the following situation when a charge particle with charge q is thrown perpendicularly to the region containing the magnetic field (inside the plane).

Magnetic field

If $L = \frac{mv}{qB}$, then how will the charge go?
I know that radius of the circular path is $\frac{mv}{qB}$, and since both are equal so, when the charge will be parallel to region(velocity along + y axis only), force acting on the particle will be in the left direction(- x axis) and since it is just on the verge on leaving the field but the force would move it inside the region and the charge particle should come out of the same face through which it entered.
But my teacher told that the particle will get deflected by 90 degree and will not come out of the face through which it entered.
Why so?

Shekhar Dangi
  • 265
  • 1
    I also got a little confused but a possible reason might be that the particle is also doing a motion in straight line along with circular motion so when it become parallel to y axis it is grazing the region of magnetic field(but not experience magnetic field) hence it will continue to move parallel to y axis It's like in TIR that the light grazes the surface of medium – Shardul Mar 04 '23 at 05:15
  • Might be the reason – Shekhar Dangi Mar 04 '23 at 06:15
  • What is the magnetic field on the rightmost boundary? If the field is zero there, the particle will follow the path your teacher described. If the field is still directed into the page on the boundary, the particle will follow the path you described. – Aiden Mar 04 '23 at 06:51

1 Answers1

1

Why so?
It all depends on how you consider the interaction between the charged particle and the right-hand "edge" of the magnetic field.

With no right-hand edge to the magnetic field present the trajectory of the charged particle will be a semicircle ending up on the left-hand edge of the magnetic field.

With the right-hand edge present the charged particle arrives "at" the right-hand edge with a velocity which is parallel to the right hand edge.
You are assuming that the particle stays in the magnetic field and carries on to complete its semicircular path but your teacher is saying that the charged particle strays outside the right hand edge and so is not subjected to any magnetic force and so carries on at a constant velocity along the edge of the magnetic field.

All purely hypothetical because there cannot be a sharp edge (instantaneous change from a finite value to zero) to a magnetic field.

Farcher
  • 95,680