91Ó°ÊÓ

A current-carrying wire is essential in many electrical and magnetic phenomena. In our scenario, the wire is placed along the x-axis and carries a steady current of 7.0 A towards the negative x direction. You'll often see wires like this in applications ranging from simple circuits to large-scale power transmission systems.

The current, a flow of electrical charge, is crucial in understanding how wires interact with magnetic fields. The direction of the current, in this case specified as negative along the x-axis, impacts how the wire experiences forces when in a magnetic field. This phenomenon is a fundamental principle of electromagnetism.

The magnetic field in our problem is vectorially described as \[\vec{B}=3.0 \hat{i}+8.0 x^{2} \hat{j},\]measured in milliteslas. This field is non-uniform, meaning it changes with respect to the x-position, as seen by the term involving \(x^2\).

Magnetic fields exert forces on moving charges or current-carrying wires. In physics, the direction and magnitude of the magnetic field directly influence the magnitude and direction of the force experienced by the wire. A constant like 3.0 along \(\hat{i}\) indicates a steady component of the field, while 8.0 \(x^2\) \(\hat{j}\) shows variability along the segment of the wire from 1.0 m to 3.0 m. Understanding this field description helps predict and calculate the resulting forces on current-carrying wires.

Calculating the force on a current-carrying wire in a magnetic field involves a cross product, denoted by \(\times\). This operation determines the direction and magnitude of the force vector, which is perpendicular to both the current direction and the magnetic field. In mathematical terms, the force \(\vec{F}\) is given by:

\[\vec{F} = I \int \vec{B} \times d\vec{l}.\]For the wire along the x-axis, the differential length \(d\vec{l} = dx \hat{i}\). Consequently, calculating \(\vec{B} \times (dx \hat{i})\) reduces terms involving \(\hat{i}\) itself to zero because \(\hat{i} \times \hat{i} = 0\).

The productive part arises from \(\hat{j} \times \hat{i} = -\hat{k}\), yielding the force along the \(\hat{k}\) direction, perpendicular to both initial vectors. This formulaic method efficiently addresses vector interactions within magnetic fields.

The definite integral is key in transitioning from the conceptual to the calculable. We begin with:
\[\int_{1}^{3} x^{2} dx.\]This calculation finds the net effect of the magnetic field over a specified segment of wire. In solving this integral, we derive:

\[\int_{1}^{3} x^{2} dx = \left[\frac{x^3}{3}\right]_{1}^{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3}.\]This result, \(\frac{26}{3}\), combines with our constants in previous steps to yield the force in millinewtons. Understanding definite integrals is crucial for solving any problem dealing with continuously changing quantities, especially in physics, where many such calculations are routine.