A small current element of length 2 cm lies at the origin along the x-axis. The magnetic field due to this element at the point (0.4 m, 0.3 m, 0) is \(2.4\times10^{-8}\ \mathrm{T}\). The current in the element is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5 A
The Biot-Savart law gives the field of a small element as \(dB=(\mu_0/4\pi)\,I\,dl\sin\theta/r^2\). The angle \(\theta\) is between the element (the x-axis) and the line from the element to the point. The point is 0.3 m from the x-axis.
Distance of the point, \(r=\sqrt{0.4^2+0.3^2}\ \mathrm{m}\), that is, 0.5 m.
Angle with the x-axis, \(\sin\theta=0.3/0.5\), that is, 0.6.
Element length, \(dl=2\times10^{-2}\ \mathrm{m}\).
\( \begin{aligned} I &= \frac{dB\,r^2}{(\mu_0/4\pi)\,dl\,\sin\theta} \\ &= \frac{2.4\times10^{-8}\times(0.5)^2}{10^{-7}\times2\times10^{-2}\times0.6} \\ &= \frac{6\times10^{-9}}{1.2\times10^{-9}} \\ &= 5\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Magnetic Field Due to a Current Element, Biot-Savart Law – “Let θ be the angle between dl and the displacement vector r. … The magnitude of this field is,”
\( dB=\frac{\mu_0}{4\pi}\frac{I\,dl\,\sin\theta}{r^2} \)\( \qquad [4.7(b)] \)
“… The proportionality constant in SI units has the exact value,”
\( \frac{\mu_0}{4\pi}=10^{-7}\ \mathrm{T\,m/A} \qquad [4.7(c)] \)