Question

An electron travels through a particular point in an experimental apparatus with a velocity of 0.949 \times 10^6×10 ​6 ​​ m/s and at an angle of +61.5^\circ ​∘ ​​ (clockwise) relative to the positive x-axis in the x-y plane. If the magnetic field at this point has a magnitude of 0.01 T and is directed in along the negative y-axis, what is the magnitude of the acceleration of the electron at this point?

Answer 1

Answer:

The magnitude of the acceleration of the electron at this point is $3.94\times10^{14}\ m/s^2$.

Explanation:

Given that,  

Velocity $v= 0.949\times10^{6}\ m/s$

Angle = 61.5°

Magnetic field = 0.01 T

We need to calculate the magnetic force

Using formula of magnetic force

$F=Bev\cos\theta$

Where, B = magnetic field

v = velocity

e = charge of electron

Put the value into the formula

$F=0.01\times1.6\times10^{-19}\times0.949\times10^{6}\cos61.5$

$F=3.59\times10^{-16}\ N$

We need to calculate the acceleration

Using newton's second law

$F= ma$

$a = \dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{3.59\times10^{-16}}{9.1\times10^{-31}}$

$a=3.94\times10^{14}\ m/s^2$

Hence, The magnitude of the acceleration of the electron at this point is $3.94\times10^{14}\ m/s^2$.