Question

A particle carrying a charge of $100\,\mu C$ moves at an angle of $45^\circ $ to a uniform magnetic field of induction  $7.5 \times 10^{-5} Wb/m^2$ with a speed of ${\text{4 }} \times {\text{ }}{10^5}{\text{ }}m/s$. Calculate the force acting on the particle.

Answer 1

Hint: When a charged particle moves in a magnetic field, it will experience a Lorentz force due to the interaction of the moving charge with the magnetic field. The direction of this force will be determined by the cross product of the velocity of the charged particle and the magnetic field

Formula used:
$\Rightarrow F = q(v \times B)$where $F$ is the Lorentz force acting on an object moving with a velocity $v$ that is moving through a magnetic field $B$

Complete step by step solution:
As the charged particle moves in a magnetic field, it will experience a Lorentz force which can be calculated from the formula:
$\Rightarrow F = q(v \times B)$
Using the property of the cross product, we can calculate the cross product of velocity and magnetic field as:
$\Rightarrow v \times B = vb\sin \theta $
Since the particle is moving at an angle of $45^\circ $ to the uniform magnetic fields, we can calculate the cross product as
$\Rightarrow v \times B = vb\sin 45^\circ = \dfrac{{vb}}{{\sqrt 2 }}$
On substituting the value of the cross product in the equation of Lorentz force (1), it can be calculated as
$\Rightarrow F = \dfrac{{qvB}}{{\sqrt 2 }}$
Substituting the values of $q = 100\mu C = 100 \times {10^{ - 6}}C$, $v = 4 \times {10^5}\,m/s$ and $B = 4 \times {10^5}\,Wb/{m^2}$, we get
$\Rightarrow F = \dfrac{{100 \times {{10}^{ - 6}} \times 4 \times {{10}^5} \times 7.5 \times {{10}^{ - 5}}}}{{\sqrt 2 }}$
$\therefore F = 2.12 \times {10^{ - 3}}\,N$ is the force experienced by the charged particle moving in the magnetic field.

Additional Information:
Had the velocity of the charged particle been perpendicular to the magnetic field, the trajectory of the particle would have been a circle in the plane perpendicular to the direction of the magnetic field. But since the velocity is at an angle of $45^\circ $ to the magnetic field, the trajectory will be a helix whose central axis will be parallel to the direction of the magnetic field.

Note:
For a charged particle moving in an electric and/or magnetic field, it will experience a Lorentz force. We must be careful in taking the angle between the velocity and magnetic field into consideration since it will affect the force acting on the charged particle.