Question

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

Short answer

The magnetic field on the Y-axis at a distance of 0.5 meters is \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\).

Step-by-step solution

Biot-Savart Law

The Biot-Savart Law states that the magnetic field (dB) created by a small element \(d\ell\), carrying a current I, can be found by the following formula: \[dB= \frac{\mu_0}{4\pi}\cdot \frac{I \cdot d\ell \times \hat{r}}{r^2}\] where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} Tm/A\)), \(d\ell\) is the length of the element, \(\hat{r}\) is the unit vector pointing from the element to the point of interest, and r is the distance from the element to the point of interest.

Set up the Biot-Savart Integral

Since we have a small element \(d\ell^-\) at the origin, the integral for finding the magnetic field at a point on the Y-axis can be set up as: \[B = \int dB = \int \frac{\mu_0}{4\pi}\cdot \frac{I \cdot d\ell \times \hat{r}}{r^2}\]

Calculate the vector cross product

Since \(d\ell\)= dx⇑ and the point P is on the positive Y-axis, the unit vector \(\hat{r}\) is in the positive Y direction as well. Evaluating the cross product dℓ × \(\hat{r}\) gives: \(d\ell \times \hat{r}\) = (dx⇑) × (0,1,0) = (0,0,-dx)

Determine the distance

Since the point P is on the positive Y-axis at a distance of 0.5 meters (50 cm) from the origin: \(r = 50\ cm\)

Evaluate the Biot-Savart integral

Now, we plug the values of the cross product and distance back into the integral: \[B = \int dB = \int \frac{\mu_0}{4\pi}\cdot \frac{I \cdot (0,0,-dx)}{50^2}\] \[B_z = -\int_{0}^{1} \frac{\mu_0 \cdot I \cdot dx}{4\pi \cdot 50^2}\] Finishing the integral, we get: \[B_z = - \frac{\mu_0 I}{4\pi \cdot 50^2}\] Now, plug in the values for \(\mu_0\) and I: \[B_z = - \frac{4\pi \times 10^{-7} \: Tm/A \cdot 10 \: A}{4\pi \cdot 50^2}\]

Solve for the magnetic field

Simplifying and solving for the magnetic field, we get: \[B_z = -2 \times 10^{-8} \: T\] Therefore, the correct answer is (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\).