A long straight solid wire of radius 2 cm carries a steady current of 10 A. The current is spread uniformly over its cross-section. The magnetic field at a point 1 cm from the axis of the wire is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(5\times10^{-5}\ \mathrm{T}\)
The point is inside the wire (\( r\lt a \)). Ampere’s law around a circle of radius \( r \) uses only the current inside that circle. For a uniform current, this enclosed current is \( I_e=Ir^{2}/a^{2} \). This gives \( B=\mu_0Ir/2\pi a^{2} \).
Here \( I=10\ \mathrm{A} \), \( a=2\ \mathrm{cm}=0.02\ \mathrm{m} \) and \( r=1\ \mathrm{cm}=0.01\ \mathrm{m} \).
\( \begin{aligned} B &= \frac{\mu_0 I r}{2\pi a^{2}} \\ &= \frac{4\pi\times10^{-7}\times10\times0.01}{2\pi\times(0.02)^{2}} \\ &= 5\times10^{-5}\ \mathrm{T} \end{aligned} \)
NCERT (2026-27):
Example 4.7 – “Now the current enclosed Ie is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is,”
\( I_e=I\left(\frac{\pi r^{2}}{\pi a^{2}}\right)=\frac{Ir^{2}}{a^{2}} \)
\( B=\left(\frac{\mu_0 I}{2\pi a^{2}}\right)r \)\( \qquad (r\lt a) \)