7. (A) A particle of charge-480 x 10-19 C passes undeflected through a velocity

Question

7. (A) A particle of charge-480 x 10-19 C passes undeflected through a velocity selector and then it penctrates perpendicularly into a uniform magnetic field. The particle hitsa detector at a distance d-0.500 m from where it entered the magnetic field as shown in the figure below. If the magnitude of the magnetic field in both the selector and deflector is B 0.400 T and if they both have the same direction then 0500 m Velocity Selector ) Determine the direction of the magnctic field B in the deflector and selector and of the electric field E in the selector and clearly indicate these directions on the figure above i) Find the speed of the particle if the magnitude of the electric field in the selector is 8.00 × NC. ili) Find the mass of the particle iv) If the directions of B and Ein the velocity selecter enly were to be reversed explain what would happen in the velocity selector.

Explanation / Answer

Hi,

7.A. In order to solve the problem, we have to remember certain things about the magnetic force as well as some definitions from this particular application of the magnetic and electric fields.

F = q ( v x B ) (1); where q is the charge of the particle, v is the speed of the particle and B is the magnetic field

Note: the direction of the magnetic force comes from using the Right Hand's Rule, but of course the sign of the charge plays an important role.

v = E/B (2) ; where B is the magnetic field and E is the electric field

m = q ( dB/2v ) (3) ; where m is the mass of the particle and d is the diameter of the circle shown in the figure.

i) The direction of the magnetic and electric fields can be determined thanks to the trajectory of the particle once is out of the velocity selector.

The magnetic force points towards the center of the semicircle shown in the figure. Using the Right Hand's Rule we can deduce that the magnetic field should go towards the outside of the page, but as the particle is negative, the direction should be the other way around. Therefore, if we use a cartesian system of coordinates we have that:

B is over the negative part of the z axis ( -k direction) :   B = B (-k)

As the direction of the magnetic force is in the i direction (over the positive part of the x axis), the electric force should be opposite to this one. The electric field should have the same direction than the electric force, but as the charge is negative, the electric field goes in the other direction. As a result:

E is over the positive part of the x axis ( i direction) :   E = E (i)

ii) The speed can be calculated using equation (2):

v = E/B = 8*104 N/C / (0.4 T) = 2*105 m/s

iii) The mass of the particle can be found using equation (3):

m = q (dB/2v) = 4.8*10-19 C [ (0.5 m)(0.4 T) / (2*2*105 m/s) ] = 2.4*10-25 kg

Note: if we consider that the mass of a proton is 1.67*10-27 kg, the mass calculated is not too strange.

iv) In this case:

B = B (k) ; E = E (-i)

This means that:

Fe = Fe (i) (electric force)

FB= FB (-i) (magnetic force)

Therefore, the forces are still being cancelled and the particle goes straight in the direction j (upper part of the y axis). Besides, when the particle is outside the velocity selector, it describes a similar trajectory that the one shown in the figure, but it will go to the left instead of going to the right.

7.B. This problem can be solved using Ampere's Law. Applying that, we have that the magnetic generated at any point by a long wire with current can be calculated by:

B = uI/(2r) (4); where u is the vacuum permeability, I is the current and r is the distance between the point of interest and the wire.

The equation (4) can calculate the module of each magnetic field, but to find the total magnetic field we have to know the direction of each one and sum the components of them.

The direction of the fields are:

B1 is over the y axis, in the lower part (angle = 270°)

B2 forms an angle of 135° with the horizontal

B3 is over the x axis, in the left part (angle = 180°)

The module of each field is equal to:

B1 = (4*10-7 Tm/A) (15 A) / [2*8*10-2 m] = 3.75*10-5 T

B2 = (4*10-7 Tm/A) (12 A) / [2*8.94*10-2 m] = 2.69*10-5 T

B3 = (4*10-7 Tm/A) (10 A) / [2*4*10-2 m] = 5*10-5 T

The components of each field is:

B1x = 0 ; B1y = - 3.75*10-5 T

B2x = cos(150°) (2.69*10-5 T) = -1.90*10-5 T ; B2y = sin(150°) (2.69*10-5 T) = 1.90*10-5 T

B3x = - 5*10-5 T ; B3y = 0

The total magnetic field is:

Bx = - 5*10-5 T -1.90*10-5 T = -6.9*10-5 T ; By = - 3.75*10-5 T + 1.90*10-5 T = -1.85*10-5 T

I hope it helps.

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