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

Correct Answer: 35

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:

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