A long wire with a small current element of length 1 cm is placed at the origin and carries a current of 10 A along the x-axis. The magnitude of the magnetic f ield, due to the element, on the y-axis at a distance 0.5 m from it, would be

The correct answer is 4 x 10⁻⁸ T.

Underlying Principle: Biot-Savart Law
The magnetic field $dB$ produced by a small current-carrying element of length $dl$ carrying a current $I$ at a point located at a distance $r$ is given by the Biot-Savart Law:

$$dB=\frac{{\mu }_0}{4\pi }·\frac{I·dl·\sin \left(\theta \right)}{r^2}$$

where:
• ${\mu }_0$ is the permeability of free space, and the constant factor $\frac{{\mu }_0}{4\pi }={10}^{-7}\text{ T·m/A}$.
• $I$ is the current flowing through the wire.
• $dl$ is the length of the current element.
• $r$ is the distance from the current element to the point where the magnetic field is measured.
• $\theta$ is the angle between the direction of the current flow (along the current element) and the position vector pointing to the observation point.

Step-by-Step Derivation and Calculation:

Step 1: Identify the given values from the problem statement
• Current, $I=10\text{ A}$
• Length of the element, $dl=1\text{ cm}={10}^{-2}\text{ m}$
• Distance, $r=0.5\text{ m}$

Step 2: Determine the angle $\theta$
The current element is placed at the origin and carries current along the positive x-axis. The point where we need to find the magnetic field lies on the y-axis. The angle between the positive x-axis (direction of current) and the positive y-axis (direction of the position vector) is ${90}^{ˆ}$.
Therefore, $\theta ={90}^{ˆ}$, which gives $\sin \left({90}^{ˆ}\right)=1$.

Step 3: Substitute the values into the formula

$$dB={10}^{-7}·\frac{10·{10}^{-2}·1}{{\left(0.5\right)}^2}$$

Simplify the numerator:
$10·{10}^{-2}=0.1$

Simplify the denominator:
${\left(0.5\right)}^2=0.25$

Now, compute the ratio:
$$dB={10}^{-7}·\frac{0.1}{0.25}$$

$$dB={10}^{-7}·0.4$$

Expressing this in scientific notation:
$$dB=4\times {10}^{-8}\text{ T}$$

Thus, the magnitude of the magnetic field due to the element at the given distance is indeed $4\times {10}^{-8}\text{ T}$.