Question

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?

Answer 1

To solve this problem it is necessary to apply the concepts related to the magnetic field and the magnetic force.

By definition we know that the magnetic force is given by

$F=q(\vec{v}x\vec{B})$

Where,

v = velocity

B= magnetic field

q= charge of a Proton $(1.6*10^{-19}C)$

We can write this equation as function of the angle between the velocity and magnetic field, i.e,

$F=qvBsin\theta$

PART A) Using Newton's second law, we can write that

$F=ma$

Equation both equation:

$ma = qvBsin\theta$

The planes between velocity and magnetic field are perpendicular, then

$ma = qvB$

$a = \frac{qvB}{m}$

Replacing

$a = \frac{(1.6*10^{-19})(2.5*10^6)(7.4*10^{-2})}{9.11*10^{-31}}$

$a = 3.25*10^{16}m/s^2$

Therefore the magnitude of maximum and minimum acceleration is $3.25*10^{16}m/s^2$

PART B) We can now calculate the angle between the electron velocity and the magnetic field through:

$a' = \frac{qvBsin\theta'}{m}$

According the statment the electrone is one-fourth of the largest magnitude of acceleration in a. Then:

$\frac{3.25*10^{16}}{4} = (3.25*10^{16})sin\theta'$

$sin\theta' = \frac{1}{4}$

$\theta' = sin^{-1}\frac{1}{4}$

$\theta' = 14.47\°$

Therefore the angle between the electron velocity and the magnetic field is 14.47°

Answer 2

(a) The maximum possible acceleration of the electron in the field is 3.16 x 10¹⁶ m/s².

(b) The angle between the electron velocity and the magnetic field is 14.5⁰.

Acceleration of the electron

The maximum possible acceleration of the electron in the field is determined by using the following formula,

ma = qvB

a = qvB/m

$a = \frac{qvB}{m} \\\\a = \frac{1.6 \times 10^{-19} \times2.5 \times 10^6 \times 7.2 \times 10^{-2} }{9.11 \times 10^{-31}} \\\\a = 3.16\times 10^{16} \ m/s^2$

Angle between the electron velocity and the magnetic field

When the actual acceleration is one-fourth of the maximum possible acceleration.

$\frac{a}{4} = \frac{qvB \times sin(\theta)}{m} \\\\\frac{3.16 \times 10^{16} }{4} = (3.16 \times 10^{16} )sin\theta\\\\\frac{1}{4} = sin(\theta)\\\\\theta = sin^{-1} (0.25)\\\\\theta = 14.5\ ^0$

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