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An electron moves at 2.50*106 m/s through a region inwhich there is a magnetic field of...

An electron moves at 2.50*106 m/s through a region inwhich there is a magnetic field of unspecified direction andmagnitude 7.40*10-2 T. (a) What are the largest andsmallest possible magnitudes of the acceleration of the electrondue to the magnetic field? (b) If the actual acceleration of theelectron is one-fourth of the largest magnitude in part (a), whatis the angle between the electron velocity and the magneticfield?
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Answer #1
Concepts and reason

The concepts required to solve the given problem are magnetic field and magnetic force.

First calculate the largest and smallest possible magnitudes of the acceleration of the electrons due to the magnetic field and then the angle between the electron velocity and the magnetic field can be calculated.

Fundamentals

Magnetic field is defined as the region around the permanent magnet where the effects of magnetic force can be experienced. Magnetic fields are produced by moving charges or current flowing in a conductor.

The SI unit of magnetic field is Tesla.

The magnitude of magnetic field is calculated by using the magnetic parts of the Lorentz force, which is given as follows:

#magnetice = qlüxB)

Here, q is the charge, v is the velocity of the moving charge, and B is the magnitude of magnetic field.

(a)

The magnetic force exists on the electron is,

#magnetice = qlüxB)

Here, q is the charge, v is the velocity of the moving charge, and B is the magnitude of magnetic field.

Rewrite the above expression as follows:

magnetic = qvBsin 0

Here, is the angle between velocity vector and magnetic field.

Substitute 90°
for and mamax
for magnetic
in the expression as follows:

F =9(VB)

UU
(916) = strany

Here, amax is the maximum acceleration and m is the mass of electron. Maximum acceleration occurs when the force acting on the charge particle is maximum. When value of sin θ
is equal to 1.

Substitute (3 6-019*1)
for q, (2.50x109 m/s)
for v, (7.40x10? T)
for B, and (8418-01*16)
for m in expression as follows:

qvB
amax
m
((1.6x10-1° C)(2.50x10^ m/s)(7.40x10^2 T)).
9.11x10-” kg
= 3.25 x 106 m/s?

(b)

The magnitude of new acceleration is,

al-qvB sin e

Here, q is the charge, v is the velocity of the moving charge, m is the mass of electron, a’ is the actual acceleration, is the angle between velocity vector and magnetic field, and B is the magnitude of magnetic field

Substitute (3.25x10 m/s
4
for and 3.25x10
m/s
for qvB
in above expression:

a=qvB sin 
(3.25*10* m52) = (325–10“ mxs*)sino
sino =
O= 14.47°

Ans: Part a

The magnitude of maximum and minimum acceleration is (3.25x10 m/s)
and 0 m/s2.

Part b

The angle between the electron velocity and the magnetic field is 14.479
.

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