A small current element produces a magnetic field of magnitude \(B\) at a point P. The point P is at a distance \(r\) from the element, on a line perpendicular to it. Another point Q is at a distance \(2r\) from the element, on a line making an angle of \(30^\circ\) with it. The magnitude of the field due to the element at Q is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(B/8\)
By the Biot-Savart law, the field of a current element varies as \(\sin\theta/r^2\). Here \(\theta\) is the angle between the element and the line joining it to the point. For P, \(\theta=90^\circ\). For Q, \(\theta=30^\circ\) and the distance is \(2r\).
\( \begin{aligned} \frac{B_Q}{B} &= \frac{\sin30^\circ/(2r)^2}{\sin90^\circ/r^2} \\ &= \frac{1}{2}\times\frac{1}{4} \\ &= \frac{1}{8} \end{aligned} \)
Hence, the field at Q is \(B/8\).
NCERT (2026-27):
Magnetic Field Due to a Current Element, Biot-Savart Law – “According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional to the current I, the element length |dl|, and inversely proportional to the square of the distance r. … The magnitude of this field is,”
\( dB=\frac{\mu_0}{4\pi}\frac{I\,dl\,\sin\theta}{r^2} \)\( \qquad [4.7(b)] \)