A 2.0 cm segment of a wire, centered at the origin (0, 0, 0) lies along X-axis. It carries a current of 4.0 A in positive X-direction. The magnetic field due to this segment at a point (0, 4.0 m, 0) is [(\(\frac{\mu_0}{4\pi}\)) = 10^-7 Tm/A, and i, j and k are unit vectors along X-axis, Y-axis and Z-axis, respectively]:

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AAI ATC Junior Executive 27 July 2022 Shift 1 Official Paper

Answer 1

Concept:

Biot Savart Law:
- Biot-savart’s law gives the magnetic field produced due to the current carrying segment.
- This segment is taken as a vector quantity known as the current element.

The magnitude of the magnetic field dB at a distance r from a current-carrying element dl is found to be proportional to I and the length dl.
Formula, \(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)
Here, \(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Calculation:

Given,

The length of segment, dl = 2.0i cm = 0.02i m

Current, I = 4 A

Position vector, \(\vec r= 4.0 \hat j m\)

\(\frac{\mu_0}{4\pi}= 10^{-7} Tm/A\)

Biot- savart law in vector form,

\(B=\frac{\mu_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)

\(B=10^{-7} ×4× \frac{0.02\hat i× 4\hat j}{4^3}\)

B = 5.0 × 10^-10k T

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