Question

The horizontal component of earth’s magnetic field at a place is $3.5 \times 10^{-5} \, \text{T}$. A very long straight conductor carrying current of $\sqrt{2} \, \text{A}$ in the direction from South East to North West is placed. The force per unit length experienced by the conductor is $\ldots \times 10^{-6} \, \text{N/m}$.

Answer 1

Calculate Magnetic Force per Unit Length: The force per unit length $\frac{F}{\ell}$ on a current-carrying conductor in a magnetic field is given by:

$\frac{F}{\ell} = i B \sin \theta$

where:

$i = \sqrt{2} \, \text{A}$ (current in the conductor)

$B = 3.5 \times 10^{-5} \, \text{T}$ (magnetic field)

$\theta = 45^\circ$ (angle between current direction and magnetic field)

Substitute Values: Using $\sin 45^\circ = \frac{1}{\sqrt{2}}$:

$\frac{F}{\ell} = (\sqrt{2}) \times (3.5 \times 10^{-5}) \times \frac{1}{\sqrt{2}} = 35 \times 10^{-6} \, \text{N/m}$

Conclusion: The force per unit length experienced by the conductor is:

$35 \times 10^{-6} \, \text{N/m}$