A cable made of several long straight wires runs in a straight line. On a circle of radius 5 cm centred on the cable, the magnetic field is tangential to the circle. Its magnitude is \(4\times10^{-5}\ \mathrm{T}\) at every point of the circle. The net current carried by the cable 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) → 10 A
On a circle centred on the cable, the field is tangential and has the same size everywhere. Ampere’s law then takes the simple form \( BL=\mu_0I_e \). Here \( L=2\pi r \) is the length of the circle and \( I_e \) is the net current through it.
Given:
\( r=5\ \mathrm{cm}=0.05\ \mathrm{m} \)
\( B=4\times10^{-5}\ \mathrm{T} \)
\( \begin{aligned} I_e &= \frac{B(2\pi r)}{\mu_0} \\ &= \frac{4\times10^{-5}\times2\pi\times0.05}{4\pi\times10^{-7}} \\ &= \frac{4\pi\times10^{-6}}{4\pi\times10^{-7}} \\ &= 10\ \mathrm{A} \end{aligned} \)
Therefore, the net current carried by the cable is 10 A. The law gives only this net value. It cannot give the current in each wire.
NCERT (2026-27):
Ampere’s Circuital Law – “… Now, let L be the length (part) of the loop for which B … is tangential. Let Ie be the current enclosed by the loop.”
\( \displaystyle BL=\mu_0 I_e \)\( \qquad [4.13(b)] \)