Question

If the magnitude of magnetic force per unit length on a wire carrying a current of 3 A and making an angle of 45° with the direction of a uniform magnetic field of $\frac{\sqrt{2}}{3}$ is $F$ N/m. Find $F$.

Answer 1

$1$

Answer 2

The magnitude of the magnetic force per unit length on a current-carrying wire in a uniform magnetic field is given by the formula: $\frac{F}{L} = I B \sin\theta$ where $I$ is the current, $B$ is the magnetic field strength, and $\theta$ is the angle between the direction of the current and the magnetic field.

Given values: Current, $I = 3$ A Magnetic field strength, $B = \frac{\sqrt{2}}{3}$ T Angle, $\theta = 45^\circ$

Substituting these values into the formula: $\frac{F}{L} = (3 \, \text{A}) \times \left(\frac{\sqrt{2}}{3} \, \text{T}\right) \times \sin(45^\circ)$ We know that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$. $\frac{F}{L} = 3 \times \frac{\sqrt{2}}{3} \times \frac{1}{\sqrt{2}}$ $\frac{F}{L} = \sqrt{2} \times \frac{1}{\sqrt{2}}$ $\frac{F}{L} = 1 \, \text{N/m}$ The question states that the magnitude of magnetic force per unit length is $F$ N/m. Therefore, $F = 1$.