An electricity and magnetism problem by Prashant Kr
A neutral atom of atomic mass number 100 which is stationary at the origin in gravity free space emits an alpha particle (A) in Z- direction. The product ion is P . A uniform magnetic field exists in the X-direction. Disregard the electro magnetic interaction between A and P . If the angle of rotation of A after which A and P meets for the first time is radians, then what is the value of n?
The answer is 48.
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1 solution
The force acting on each particles is given by F = q v × B .
Since both particles are moving perpendicular to the field, this simplifies to q v B
The charge on both particles is the same (in magnitude) and the field is the same. Since we are only interested in relative speed of motion, we will set both q and B to 1 for convenience.
So both particles have a force acting on them equal to v, their velocity.
An alpha particle has a mass of 4, so P has a mass of 96
By momentum conservation, v A = 2 4 v P because P is 24 times as massive as A. We'll just say :
v A = 2 4 and v P = 1
Therefore F A = 2 4 and F P = 1
a = m F . m A = 1 ; m B = 2 4
Therefore accelerations: a A = 2 4 a P = 2 4 1
These particles will move in circles with radius a v 2
This gives both particles travelling in circles of equal radius.
Since the particles have opposite charges, one will turn right and one left. Since they are moving in opposite directions, this means that they are tracing opposite paths around the same circle. Since, from earlier, A is moving 24 x faster than P, they will meet when A is 24/25 of the way around the circle. Therefore n = 48.