91Ó°ÊÓ

We can use the formula for the magnetic field generated by a long straight wire carrying a current \(I\) at a distance \(r\) away from the wire: \[B = \frac{\mu_0 I}{2 \pi r}\] Where \(B\) is the magnetic field, \(\mu_{0}\) is the magnetic constant, \(I\) is the current, and \(r\) is the distance from the wire. Since the point charge is equidistant from the two wires, it is located at a distance \(r = d/2\) from each wire. For wire carrying current \(I\), the magnetic field at the location of the charge will be: \[B_{1} = \frac{\mu_{0} I}{2 \pi (d/2)} = \frac{\mu_{0} I}{\pi d}\] For wire carrying current \(-I\), the magnetic field at the location of the charge will be: \[B_{2} = -\frac{\mu_{0} I}{2 \pi (d/2)} = -\frac{\mu_{0} I}{\pi d}\]

Since the magnetic fields from each wire have opposite directions but equal magnitudes, they cancel each other out at the location of the charge: \[B_{net} = B_{1} + B_{2} = \frac{\mu_{0} I}{\pi d} - \frac{\mu_{0} I}{\pi d} = 0\]

To calculate the force on the charge, we use the Lorentz force formula: \[F = q(\vec{v} \times \vec{B})\] Since the magnetic field is \(0\) at the location of the charge, the force on the charge will also be \(0\): \[F = q(\vec{v} \times \vec{B_{net}}) = q(\vec{v} \times 0) = 0\]

The magnitude of the force due to the magnetic field acting on the charge at this instant is: (D) zero