Question

An electron is moving through a magnetic field whose magnitude is 83 x 10-5 T. The electron experiences only a magnetic force and has an acceleration of magnitude 34 x 10+13 m/s2. At a certain instant, it has a speed of 72 x 10+5 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.

Answer 1

Answer:

the angle between the electron's velocity and the magnetic field is 19⁰

Explanation:

Given;

magnitude of the magnetic field, B = 83 x 10⁻⁵ T

acceleration of the electron, a = 34 x 10¹³ m/s²

speed of the electron, v = 72 x 10⁵ m/s

mass of electron, m = 9.11 x 10⁻³¹ kg

The magnetic force experienced by the electron is calculated as;

F = ma = qvB sinθ

where;

q is charge of electron = 1.602 x 10⁻¹⁹ C

θ is the angle between the electron's velocity and the magnetic field.

$sin(\theta ) = \frac{ma}{qvB} \\\\sin(\theta ) = \frac{(9.11\times 10^{-31})(34\times 10^{13})}{(1.602\times 10^{-19})\times (72\times 10^5) \times (83 \times 10^{-5})} \\\\sin(\theta ) = 0.3235\\\\\theta =sin^{-1}(0.3235)\\\\\theta =18.9^0$

$\theta \approx 19^ 0$

Therefore, the angle between the electron's velocity and the magnetic field is 19⁰