Question

An electron is moving through a magnetic field whose magnitude is 9.21 × 10^-4 T. The electron experiences only a magnetic force and has an acceleration of magnitude 2.30 × 10^14 m/s^2. At a certain instant, it has a speed of 7.69 × 10^6 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.

Answer 1

Answer:

$\theta=10.60^{\circ}$

Explanation:

Given that,

Magnetic field, $B=9.21\times 10^{-4}\ T$

Acceleration of the electron, $a=2.3\times 10^{14}\ m/s^2$

Speed of electron, $v=7.69\times 10^{6}\ m/s$

The force due to this acceleration is balanced by the magnetic force as :

$ma=qvB\ sin\theta$

$sin\theta=\dfrac{ma}{qvB}$

$sin\theta=\dfrac{9.1\times 10^{-31}\times 2.3\times 10^{14}}{1.6\times 10^{-19}\times 7.69\times 10^{6}\times 9.21\times 10^{-4}}$

$\theta=10.60^{\circ}$

So, the angle between the electron's velocity and the magnetic field is 10.6 degrees. Hence, this is the required solution,