10.4 Motion of a Charged Particle in a Magnetic Field

10.4 Motion of a Charged Particle in a Magnetic Field

We have observed that a current-carrying conductor placed perpendicularly in a uniform magnetic field experiences a force $F$. Since a current is the flow of electric charges, it raises the question: do individual charges moving through a magnetic field also experience a force? The answer is yes. Experiments show that a charged particle does experience a force when it moves across a magnetic field. We can calculate this force by examining the behaviour of a current-carrying conductor in a magnetic field.

Consider a conductor of length $L$ through which $N$ charged particles, each with charge $q$, are passing in time $t$. The motion of these charged particles produces a current $I$ in the conductor, which is given by:

where $Q$ is the total charge flowing in time $t$. If $v$ is the velocity of charged particles, then the velocity of the particles along the conductor is:

where $\hat{L}$ is the unit vector in the direction of the current. The sign of the force depends on whether the charge $q$ is positive or negative. However, the unit vector $\hat{L}$ is directed along the direction of the current, which is the direction of motion of positive charges.

Since the particles take time $t$ to move across the conductor of length $L$, therefore:

Then

If this conductor is placed in a uniform magnetic field $B$, it will experience a force $F$, as given by Eq. 10.3:

As $I = (Nqv/L)\hat{L}$ $\ldots (10.10)$

Substituting the values of $I$ and $L$ in Eq. (10.3), we have:

or

As velocity $v$ is directed along the conductor $\hat{L}$, so we can write:

$\mathbf{v}\hat{L} = v\hat{L} = \mathbf{v}$ (as $\hat{L} = \hat{v}$)

Therefore the force on a single particle will be:

If $\theta$ is the angle between $\mathbf{B}$ and $\mathbf{v}$, then magnitude of force $F$ is given by:

Therefore, the force is maximum when $B$ is perpendicular to $v$, i.e., $\theta = 90^{\circ}$, and force is zero when $B$ is in the direction of $v$, i.e., $\theta = 0$. The direction of force can be known by applying Fleming's left hand rule or right hand rule for vector product.

(a) The positively charged particle enters into the magnetic field along the dotted line on plane of paper. It experiences a force in the upward direction due to which it is deflected along a curved path (Fig. 10.10-a).

(b) The negatively charged particle is deflected downward by the force acting on it downwards (Fig. 10.10-b).

Example 10.3

An electron enters into a uniform magnetic field perpendicularly with a speed of $10^{4}\,\mathrm{m\,s^{-1}}$. What path the electron will move along inside the field?

$B = 2.5\,\mathrm{Wb\,m^{-2}}$, $m = 9.11 \times 10^{-31}\,\mathrm{kg}$, $e = 1.6 \times 10^{-19}\,\mathrm{C}$

Solution

The Force Acting on the electron will be:

As the velocity $v$ is perpendicular to $B$, i.e., $\theta = 90^{\circ}$, the charge $q = e$, so $F = evB \times 1 = evB$

Since $F$ acts perpendicular to $v$, so this force provides centripetal force to the electron to keep it in a circle of radius $r$. Then:

Putting the values in above equation, we have:

The path of electron will be a circle of radius $2.3 \times 10^{-8}\,\mathrm{m}$.