A circular loop of radius R and a long straight wire carry equal steady currents. The ratio of the magnitude of the magnetic field at the centre of the loop to that at a distance R from the wire is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \pi : 1 \)
Both fields come from the Biot-Savart law. At the centre of a circular loop, every element is at the same distance R. Each element is also at right angles to the line joining it to the centre. This gives \( \mu_0 I/2R \). For a long straight wire the field falls off as \( 1/r \).
The field at the centre of the loop is:
\( \displaystyle B_{\mathrm{loop}}=\frac{\mu_0 I}{2R} \)
The field at a distance R from the wire is:
\( \displaystyle B_{\mathrm{wire}}=\frac{\mu_0 I}{2\pi R} \)
Dividing the two:
\( \begin{aligned} \frac{B_{\mathrm{loop}}}{B_{\mathrm{wire}}} &= \frac{\mu_0 I/2R}{\mu_0 I/2\pi R} \\ &= \pi \end{aligned} \)
Therefore, the ratio is \( \pi : 1 \). The loop gives the stronger field.