A 10 -cm long portion of a straight wire carrying a 10 A current is...

The magnetic force on a current-carrying wire is given by the formula:

$\vec{F} = I(\vec{L} \times \vec{B})$

The magnitude of the force is:

$F = I L B \sin\theta$

Given values are:

$I = 10 \text{ A}$

$L = 10 \text{ cm} = 0.1 \text{ m}$

$B = 0.1 \text{ T}$

\theta = 53^{}^{{}^{\circ}}

Substituting these values into the formula:

F = 10 \times 0.1 \times 0.1 \times \sin(53^{}^{{}^{\circ}})

$F = 0.1 \times \dfrac{4}{5} = 0.08 \text{ N}$

From the cross product $\vec{L} \times \vec{B}$, the direction of the magnetic force is perpendicular to both the wire and the magnetic field.