Question

An element \(\overrightarrow{\Delta l}=\Delta \mathrm{xi}\) is placed at the origin and carries a current of 10 A . The magnitude of magnetic field on the Y axis at a distance of 0.5 m if \(\Delta x=1 \mathrm{~cm}\) is \(\left(\frac{\mu_0}{4 \pi}=10^{-7}\right.\) SI unit \()\left(\sin 90^{\circ}=1\right)\)

• A. \(2\times 10^{-7}\text{ }\mathrm{T}\)
• B. \(10^{-8}\text{ }\mathrm{T}\)
• C. \(4\times 10^{-8}\text{ }\mathrm{T}\)
• D. \(2\times 10^{-8}\text{ }\mathrm{T}\)

Hint:

Use the Biot-Savart law with the element along X and the point on Y, so the angle is 90 degrees.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Biot-Savart law: \(dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,\Delta l\sin\theta}{r^2}\).

Step 2: Substitute
\(I=10\text{ A}\), \(\Delta l=10^{-2}\text{ m}\), \(r=0.5\text{ m}\), \(\sin90^{\circ}=1\).

Step 3: Calculation
\(B=10^{-7}\times\dfrac{10\times10^{-2}}{(0.5)^2}=10^{-7}\times\dfrac{0.1}{0.25}=4\times10^{-8}\text{ T}\).

Step 4: Compare the options
This matches the third option.

Final Answer:
The field magnitude is 4 times 10 to the power minus 8 tesla. \[ \boxed{\text{(C) }4\times 10^{-8}\text{ T}} \]