10.1 Force on a Current-Carrying Conductor in a Uniform Magnetic Field

10.1 Force on a Current-Carrying Conductor in a Uniform Magnetic Field

It has been observed experimentally that a current-carrying conductor placed in a magnetic field experiences a force. Consider a straight conductor carrying a steady current placed perpendicular to uniform magnetic field. Assume the direction of the current is out of the paper as shown by $\odot$ in Fig. 10.2. The direction of magnetic field produced by the current is also shown.

Just as two magnets exert forces on each other through their magnetic fields, a current-carrying conductor experiences a force due to the interaction between its own magnetic field and the external magnetic field. To determine the direction of this force, consider the interaction between the two fields.

The magnetic field produced by the current and the external uniform magnetic field reinforce each other on the left side of the conductor and oppose each other on the right side. Consequently, the conductor moves towards the side where the field is weaker. That is, the force on the conductor is directed to the right as viewed from the front. Thus, the force $F$ is perpendicular to both the conductor and the magnetic field.

Fleming's left-hand rule is used to predict the direction of the force experienced by a current-carrying conductor in a magnetic field. To apply the rule:

lightbulb

Position your left hand such that the first finger points in the direction of magnetic field, the second finger points in the direction of current, the thumb will then point in the direction of force.

The Fleming's left hand rule is illustrated in Fig. 10.3.

However, direction of force can also be found by using right hand rule that can be stated as:

lightbulb

Curl fingers of your right hand from current to magnetic field through smaller angle, the stretched thumb will indicate the direction of force.

Let us now determine the magnitude of the force on a current-carrying conductor placed inside a magnetic field. Experimentally, it has been observed that the magnitude of the force acting on the conductor is directly proportional to the current $I$ in the conductor, the length $L$ of the conductor, and the strength of the external magnetic field $B$. The strength of the magnetic field is also known as the magnetic induction $B$, which has the same direction as the field. Thus, the force $F$ on a conductor of length $L$, carrying a current $I$ and placed perpendicular to a magnetic field of strength $B$, is given by:

In SI units, the value of $k = 1$. Therefore:

From Eq. (10.1) we can see that $B = F/IL$, so we can define $B$ as:

lightbulb

The magnetic strength is numerically equal to the force exerted on a conductor of length one metre carrying one ampere current, placed perpendicular to the magnetic field.

Equation $B = F/IL$ also gives us the unit of $B$. The SI unit of $B$ is tesla $(\mathrm{T})$.

It may be noted that magnetic induction is a vector quantity. Its direction is the same as that of magnetic field.

We can also consider a vector $\mathbf{L}$ which has a magnitude equal to the length of the conductor and its direction is along the flow of current.

Now consider a conductor $L$ placed at an angle $\theta$ with respect to the magnetic field. Then we will use the component of $L$ perpendicular to $B$ i.e., $(L\sin\theta)$, as shown in Fig. 10.4. Then the Eq. (10.1) will become:

In the vector form the Eq. 10.2 can be written as:

Equation (10.2) shows that the force will be maximum $(BIL)$ when the conductor is perpendicular to the field, i.e., $\theta = 90^{\circ}$, and it will be zero when the conductor is along the field i.e., $\theta = 0$.

Example 10.1

A $20.0\,\mathrm{cm}$ wire carrying a current of $10.0\,\mathrm{A}$ is placed in a uniform magnetic field of $0.30\,\mathrm{T}$. If the wire makes an angle of $40^{\circ}$ with the direction of magnetic field, find the magnitude of the force acting on the wire.

Solution

Length of the wire $= L = 20.0\,\mathrm{cm} = 0.20\,\mathrm{m}$

Current $I = 10.0\,\mathrm{A}$

Strength of magnetic field $B = 0.30\,\mathrm{T}$

Angle $\theta = 40^{\circ}$

Substituting the values in Eq. (10.2):

Interpreting Information

lightbulb

The Earth's magnetic field is approximately that of a dipole, like that of the fictitious bar magnet, where the south magnetic pole is towards the geographic north pole and the north magnetic pole is towards the geographic south pole.