Question

Magnetic field at a certain place is 3.0 × $10^{-5}$ T and the direction of the field is from the south to the north. A very long straight conductor is carrying a steady current of 2A. What is the magnitude of force per unit length on it when it is placed on a horizontal table and the direction of the current is east to west?

Answer 1

6.0 × 10^{-5} N m^{-1}

Answer 2

The force per unit length on a current-carrying conductor in a magnetic field is given by the formula:

$f = \frac{F}{L} = I B \sin\theta$

where:

$f$ is the force per unit length $I$ is the current flowing through the conductor $B$ is the magnetic field strength $\theta$ is the angle between the direction of the current and the direction of the magnetic field.

Given values:

Magnetic field strength, $B = 3.0 \times 10^{-5}$ T Direction of magnetic field: South to North Current, $I = 2$ A Direction of current: East to West

The magnetic field is directed from South to North, and the current is directed from East to West. These two directions are perpendicular to each other in a horizontal plane. Therefore, the angle $\theta$ between the current direction and the magnetic field direction is $90^\circ$. So, $\sin\theta = \sin 90^\circ = 1$.

Substitute the values into the formula:

$f = (2 \text{ A}) \times (3.0 \times 10^{-5} \text{ T}) \times \sin 90^\circ$ $f = 2 \times 3.0 \times 10^{-5} \times 1$ $f = 6.0 \times 10^{-5} \text{ N m}^{-1}$

The magnitude of the force per unit length on the conductor is $6.0 \times 10^{-5}$ N m$^{-1}$.