There is a thin conducting wire carrying current. The value of magnetic field induction at any point on the conductor would be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Zero
By Ampere's circuital law, the magnetic field at a distance r from the axis of a straight conductor of radius a, for \( r \le a \), is due only to the current inside the circle of radius r:
\( \displaystyle B\,(2\pi r) = \mu_0\,I\,\frac{r^{2}}{a^{2}} \Rightarrow B = \frac{\mu_0 I\,r}{2\pi a^{2}} \)
For a thin wire the points of the conductor lie on its axis, where \( r = 0 \). The enclosed current is zero there, and by the symmetry of the current distribution the field has no preferred direction, so
\( B = 0 \)
The field is therefore not a constant (it grows with r from zero), and a value of zero is neither positive nor negative.