A current element of length 0.8 cm carrying a current of 5 A towards +x - direction is placed symmetrically at the origin along the x-axis. The magnetic field at a point (0, 5 cm) is:

(i, j, and k are unit vectors along the x-axis, y-axis and z-axis, respectively.)

Question

Question

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AAI ATC Junior Executive 27 July 2022 Shift 3 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, \(B=\frac{μ_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)
Here, \(\frac{μ_0}{4\pi}= 10^{-7} ~Tm/A\)

Calculation:

Given,

The length of segment, dl = 0.80i cm = 0.008i m

Current, I = 5 A

Position vector, \(\vec r= 5.0 \hat j cm = 0.05 \hat j m\)

\(\frac{μ_0}{4\pi}= 10^{-7} ~Tm/A\)

Biot- savart law in vector form,

\(B=\frac{μ_0I}{4\pi}\frac{d\vec l× \vec r}{|\vec r|^3}\)

\(B=10^{-7} ×5× \frac{0.008\hat i× 0.05\hat j}{(0.05)^3}\)

B = 1.6 μk T

Hence, the magnetic field is 1.6μ kT.