Question

A circular wire of radius \(5.0 \mathrm{~cm}\) has a current of \(3.0 \mathrm{~A}\) flowing in it. The wire is placed in a uniform magnetic field of \(5.0 \mathrm{mT}\). a) Determine the maximum torque on the wire. b) Determine the range of the magnetic potential energy of the wire.

Short answer

Answer: The maximum torque on the wire is approximately 0.00011775 Nm, and the range of the magnetic potential energy of the wire is from -0.00011775 J to 0.00011775 J.

Step-by-step solution

Understand the given parameters

Here, we are given the radius of the circular wire as \(5.0 \mathrm{~cm}\) or \(0.05 \mathrm{~m}\), the current in the wire as \(3.0 \mathrm{~A}\), and the magnetic field as \(5.0 \mathrm{mT}\) or \(5.0 \times 10^{-3} \mathrm{T}\).

Calculate the magnetic moment of the wire

The magnetic moment (\mu) of the circular wire can be calculated with the following formula: \(\mu = IA\), where \(I\) is the current and \(A\) is the area of the coil. First, let's calculate the area of the circular wire, which can be found using the formula: \(A = \pi r^{2}\), where r is the radius of the wire. Plug the radius value into the formula: \(A = \pi (0.05)^{2}\) \(A \approx 0.00785 \mathrm{~m^2}\) Now, we can calculate the magnetic moment: \(\mu = (3.0\,\mathrm{A})(0.00785\,\mathrm{m^2})\) \(\mu \approx 0.02355\,\mathrm{A\,m^2}\)

Calculate the maximum torque on the wire

The torque (\tau) on a current-carrying loop in a magnetic field can be calculated using the following formula: \(\tau = \mu B \sin{\theta}\), where \(\mu\) is the magnetic moment, \(B\) is the magnetic field, and \(\theta\) is the angle between \(\mu\) and \(B\). For the maximum torque, we know that \(\sin{\theta} = 1\), since the torque is maximum when \(\theta = 90^\circ\). Plug the values into the formula: \(\tau_{max} = (0.02355\,\mathrm{A\,m^2})(5.0 \times 10^{-3}\,\mathrm{T})(1)\) \(\tau_{max} \approx 0.00011775\,\mathrm{N\,m}\)

Calculate the range of the magnetic potential energy of the wire

The magnetic potential energy (U) can be calculated using the following formula: \(U = -\mu B \cos{\theta}\) The range of the magnetic potential energy can be calculated by finding the maximum and minimum values of U, which occur at \(\theta = 0^\circ\) and \(\theta = 180^\circ\), respectively. For \(\theta = 0^\circ\): \(U_{max} = -\mu B \cos{(0^\circ)}\) \(U_{max} = -0.02355 \times 5.0 \times 10^{-3} \times 1\) \(U_{max} = -0.00011775\,\mathrm{J}\) For \(\theta = 180^\circ\): \(U_{min} = -\mu B \cos{(180^\circ)}\) \(U_{min} = -0.02355 \times 5.0 \times 10^{-3} \times (-1)\) \(U_{min} = 0.00011775\,\mathrm{J}\) The range of the magnetic potential energy is from \(-0.00011775\,\mathrm{J}\) to \(0.00011775\,\mathrm{J}\). In conclusion: a) The maximum torque on the wire is \(\approx 0.00011775\,\mathrm{N\,m}\). b) The range of the magnetic potential energy of the wire is from \(-0.00011775\,\mathrm{J}\) to \(0.00011775\,\mathrm{J}\).