Question

a) A loop of wire \(5.00 \mathrm{~cm}\) in diameter is carrying a current of \(2.00 \mathrm{~A}\) What is the energy density of the magnetic field at its center? b) What current has to flow in a straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire?

Short answer

Answer: The current needed in a straight wire to produce the same energy density at a point 4.00 cm away is approximately 1.59 A.

Step-by-step solution

Calculate the magnetic field at the center of the loop

Using Ampere's law, the magnetic field at the center of a loop with radius \(r\) and current \(I\) is given by the formula: \(B = \frac{\mu_0 I}{2r}\) where \(\mu_0 = 4\pi\times10^{-7} \, T \cdot m/A\) is the permeability of free space. Given, diameter \(d = 5.00 \, cm\) and current \(I = 2.00 \, A\). The radius \(r\) of the loop is half the diameter, so \(r = \frac{d}{2} = 2.50\,cm = 0.025\,m\). Now, substitute the values and find \(B\): \(B = \frac{4\pi\times10^{-7}\cdot 2.00}{2\cdot0.025} = 1.00\times10^{-5}\,T\)

Calculate the energy density of the magnetic field

The energy density of the magnetic field can be calculated using the formula: \(u = \frac{1}{2} \frac{B^2}{\mu_0}\) Plug in the values and calculate the energy density: \(u = \frac{1}{2} \frac{(1.00\times10^{-5})^2}{4\pi\times10^{-7}} = 3.98\times10^{-4} \, J/m^3\) So, the energy density of the magnetic field at the center of the loop is \(3.98\times10^{-4} \, J/m^3\).

Calculate the magnetic field at the given point from a straight wire

We know the energy density and we need to find the magnetic field, so rearrange the energy density formula: \(B_1 = \sqrt{2u\mu_0}\) Plugging the energy density we calculated, we can find the magnetic field: \(B_1 = \sqrt{2\cdot3.98\times10^{-4}\cdot 4\pi\times10^{-7}} = 2.00\times10^{-5}\,T\)

Calculate the current in the straight wire

Using Ampere's law for an infinite straight current-carrying wire, the magnetic field B at a perpendicular distance \(r\) from a wire with current I is given by: \(B = \frac{\mu_0 I}{2\pi r}\) We have \(B = 2.00\times10^{-5}\,T\) and distance \(r = 4.00\,cm = 0.040\,m\). We need to find \(I\). Rearrange the formula to solve for \(I\): \(I = \frac{2\pi r B}{\mu_0}\) Substituting the values, we get: \(I = \frac{2\pi\cdot0.040\cdot 2.00\times10^{-5}}{4\pi\times10^{-7}} = 1.59\,A\) Therefore, the current that must flow in a straight wire to produce the same energy density at a point \(4.00\,cm\) away is approximately \(1.59\,A\).