Question

Write the expression for the magnetic field due to a current element in vector form. Consider a \(1\,\text{cm}\) segment of a wire, centered at the origin, carrying a current of \(10\,\text{A}\) in positive \(x\)-direction. Calculate the magnetic field \(\vec{B}\) at a point \((1\,\text{m},1\,\text{m},0)\).

Hint:

For numerical Biot–Savart problems: \[ d\vec B= \frac{\mu_0}{4\pi} \frac{I(d\vec l\times\vec r)}{r^3} \] is often easier to use directly because it avoids separately calculating the unit vector \(\hat r\).

Answer 1

Concept: The magnetic field due to a small current element is given by the Biot–Savart law. In vector form, \[ d\vec B = \frac{\mu_0}{4\pi} \frac{I(d\vec l\times \hat r)}{r^2}. \] Alternatively, \[ d\vec B = \frac{\mu_0}{4\pi} \frac{I(d\vec l\times \vec r)}{r^3}. \] where
• \(I\) is the current,
• \(d\vec l\) is the current element,
• \(\vec r\) is the position vector from the current element to the observation point,
• \(r\) is its magnitude.

Step 1: Write the given quantities. Current: \[ I=10\,\text{A} \] Length of current element: \[ dl=1\,\text{cm}=10^{-2}\,\text{m} \] Since current flows along positive \(x\)-direction, \[ d\vec l=(10^{-2})\hat i. \] Observation point: \[ P(1,1,0). \] Since the current element is centered at the origin, \[ \vec r = 1\hat i+1\hat j. \] Magnitude: \[ r=\sqrt{1^2+1^2} = \sqrt2\,\text{m}. \]

Step 2: Calculate the vector product \(d\vec l\times\vec r\). \[ d\vec l\times\vec r = (10^{-2}\hat i) \times (\hat i+\hat j). \] Using \[ \hat i\times\hat i=0, \] and \[ \hat i\times\hat j=\hat k, \] we get \[ d\vec l\times\vec r = 10^{-2}\hat k. \] Thus, \[ \boxed{ d\vec l\times\vec r = 10^{-2}\hat k }. \]

Step 3: Apply the Biot–Savart law. \[ d\vec B = \frac{\mu_0}{4\pi} \frac{I(d\vec l\times\vec r)}{r^3}. \] Substituting \[ \frac{\mu_0}{4\pi}=10^{-7}, \] \[ I=10, \] \[ d\vec l\times\vec r=10^{-2}\hat k, \] and \[ r^3=(\sqrt2)^3=2\sqrt2, \] we obtain \[ d\vec B = 10^{-7} \frac{10\times10^{-2}} {2\sqrt2} \hat k. \] \[ d\vec B = 10^{-8} \frac{1}{2\sqrt2} \hat k. \] \[ d\vec B = 3.54\times10^{-9}\hat k\;\text{T}. \]

Step 4: Determine the direction. The direction is along \[ \hat k \] which corresponds to the positive \(z\)-axis. This is consistent with the right-hand rule. Final Answer: Biot–Savart law: \[ \boxed{ d\vec B = \frac{\mu_0}{4\pi} \frac{I(d\vec l\times\hat r)}{r^2} } \] Magnetic field at \((1,1,0)\): \[ \boxed{ \vec B = 3.54\times10^{-9}\,\hat k\;\text{T} } \] or \[ \boxed{ \vec B = 3.54\times10^{-9}\,\text{T} \text{ along the positive }z\text{-axis} }. \]