Question

What is the magnitude of the magnetic field inside a long, straight tungsten wire of circular cross section with diameter \(2.4 \mathrm{~mm}\) and carrying a current of \(3.5 \mathrm{~A},\) at a distance of \(0.60 \mathrm{~mm}\) from its central axis?

Short answer

Answer: The magnitude of the magnetic field inside the wire at a distance of \(0.60\textrm{ mm}\) from its central axis is approximately \(5.98 \times 10^{-6}\textrm{ T}\).

Step-by-step solution

Identify the variables

First, let's identify the variables given in the problem. Diameter (D) = \(2.4\textrm{ mm} = 2.4 \times 10^{-3}\textrm{ m}\)\ Radius (R) = Diameter/2 = \(1.2\textrm{ mm} = 1.2 \times 10^{-3}\textrm{ m}\)\ Current (I) = \(3.5\textrm{ A}\)\ Distance from the central axis (r) = \(0.60\textrm{ mm} = 0.60 \times 10^{-3}\textrm{ m}\) Now we need to find the magnetic field B at a distance r from the central axis.

Use Amperes Law

We can use Ampere's Law and the formula for the magnetic field inside a wire to solve for B. Ampere's Law states that \(\oint \vec{B} \cdot d\vec{l} = \mu_{0}I_{encl}\), where\ \(\oint \vec{B} \cdot d\vec{l}\) is the integral of the magnetic field around a closed loop,\ \(\mu_{0}\) is the permeability of free space (approximately \(4\pi × 10^{−7} \textrm{ T·m / A}\)),\ \(I_{encl}\) is the current enclosed by the loop.

Calculate the current enclosed

Now, we need to find the current enclosed by the loop (I_enc). We can do this using the ratio of the areas of the wire. \(I_{encl} = I \times \frac{\pi r^{2}}{\pi R^{2}}\) Substituting the values, we get,\ \(I_{encl} = 3.5 \times \frac{\pi (0.60 \times 10^{-3})^{2}}{\pi (1.2 \times 10^{-3})^{2}}\)

Calculate the magnetic field

Now, we can rearrange Ampere's Law and solve for B. Since \(B=\frac{\oint \vec{B} \cdot d\vec{l}}{2\pi r}\),\ \(B = \frac{\mu_{0}I_{encl}}{2\pi r}\) Substituting the values, we get,\ \(B = \frac{(4\pi × 10^{−7})(3.5) \left(\frac{\pi (0.60 \times 10^{-3})^{2}}{\pi (1.2 \times 10^{-3})^{2}}\right)}{2\pi(0.60\times 10^{-3})}\) Solve for B,\ \(B \approx 5.98 \times 10^{-6} \textrm{ T}\) Therefore, the magnitude of the magnetic field inside the wire at a distance of \(0.60\textrm{ mm}\) from its central axis is approximately \(5.98 \times 10^{-6}\textrm{ T}\).