Read the following passage and answer the questions.
A student studies the magnetic field produced by a steady current. She uses the Biot-Savart law. First she takes a short straight piece of wire, 1 cm long, placed at the origin. It carries a current of 5 A in the +x direction. She marks some points, each 10 cm from this piece. Some of these points lie on the x-, y- and z-axes. One point, Q, lies in the x-y plane, on a line that makes an angle of 30° with the +x-axis. She also moves a point P along the +y-axis to see how the field changes with distance. Next, she bends a long wire into a flat circular loop of radius \(R\) and passes a steady current \(I\) through it. She treats the loop as a large number of small current elements. She then studies the field at the centre of the loop and at points on its axis.
The magnitude of the magnetic field at Q due to the 1 cm piece of wire is (Take \(\mu_0/4\pi=10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.5\times10^{-7}\ \mathrm{T}\)
By the Biot-Savart law, the field of a small current element depends on an angle \( \theta \). This is the angle between the element and the line joining it to the point. The magnitude is \( dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^{2}} \).
Given:
\( I=5\ \mathrm{A} \)
\( dl=1\ \mathrm{cm}=10^{-2}\ \mathrm{m} \)
\( r=10\ \mathrm{cm}=0.1\ \mathrm{m} \)
\( \theta=30^\circ \)
\( \begin{aligned} dB &= \frac{\mu_0}{4\pi}\,\frac{I\,dl\sin\theta}{r^{2}} \\ &= \frac{10^{-7}\times5\times10^{-2}\times0.5}{(0.1)^{2}} \\ &= 2.5\times10^{-7}\ \mathrm{T} \end{aligned} \)
Therefore, the field at Q is \( 2.5\times10^{-7}\ \mathrm{T} \), half of the value at the same distance on the y-axis.
NCERT (2026-27):
Magnetic Field Due to a Current Element, Biot-Savart Law – “The magnitude of this field is,”
\( \displaystyle dB=\frac{\mu_0}{4\pi}\,\frac{I\,dl\sin\theta}{r^{2}} \)\( \qquad [4.7(b)] \)