Question

A current of 10 A is flowing in a long wire along the positive Z-axis. Find the intensity of magnetic field at a point (10 cm, 0, 0).

Hint:

Always ensure all units are in the SI system (meters, Amperes, etc.) before performing calculations. For finding the direction of the magnetic field from a straight wire, the Right-Hand Thumb Rule is the quickest method.

Answer 1

Step 1: Understanding the Concept:
A long, straight wire carrying a current produces a magnetic field in the form of concentric circles around the wire. The magnitude of this field can be calculated using Ampere's Law or the Biot-Savart Law.

Step 2: Key Formula or Approach:
For an infinitely long, straight wire carrying a current \(I\), the magnitude of the magnetic field \(B\) at a perpendicular distance \(r\) from the wire is given by: \[ B = \frac{\mu_0 I}{2 \pi r} \] where \(\mu_0\) is the permeability of free space, with a value of \(4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}\).

Step 3: Detailed Explanation:
First, identify the given values and convert them to SI units: \begin{itemize} \item Current, \(I = 10\) A.
\item The wire is along the Z-axis.
\item The point is P(10 cm, 0, 0).
\item The perpendicular distance \(r\) of the point P from the Z-axis is its x-coordinate, which is 10 cm.
\item Convert the distance to meters: \(r = 10 \text{ cm} = 0.1 \text{ m}\).
\end{itemize} Now, substitute these values into the formula for the magnetic field: \[ B = \frac{(4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \times (10 \, \text{A})}{2 \pi \times (0.1 \, \text{m})} \] Cancel out \(2\pi\) from the numerator and denominator: \[ B = \frac{2 \times 10^{-7} \times 10}{0.1} \] \[ B = \frac{2 \times 10^{-6}}{10^{-1}} \] \[ B = 2 \times 10^{-6 - (-1)} = 2 \times 10^{-5} \, \text{T} \]

Step 4: Final Answer:
The intensity of the magnetic field at the point (10 cm, 0, 0) is \(2 \times 10^{-5}\) Tesla.
(Note: To find the direction, use the right-hand thumb rule. If the thumb points along the current (+Z direction), the curled fingers at the point (+X axis) will point in the +Y direction. So, \(\vec{B} = 2 \times 10^{-5} \hat{j} \, \text{T}\)).