Question

A long thick straight hollow conductor (tube) carries a current $I$, it has two sections $\text{A}$ and $\text{C}$ of unequal cross-sections joined by a conical section $\text{B}$ as shown in the figure. $1,2$ and $3$ are points on a line parallel to the axis of the conductor. The magnetic fields at points $1,2$ and $3$ have magnitudes $B_1,B_2$ and $B_3$ respectively, then which of the following is correct?

• A. $B_1=B_2=B_3$
• B. $B_1=B_2\ne B_3$
• C. $B_1<B_2<B_3$
• D. $B_2$ cannot be found unless the dimensions of the section B are known

Answer 1

The correct option is A $B_1=B_2=B_3$



To find the magnetic field outside a thick conductor , the current may be assumed to flow along its axis.

Using Amper's law,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0(I_{net}{)}_{enclosed}$

Therefore, the magnetic field for long straight conductor is,

$B=\frac{{\mu }_{0}i}{2\pi d}$

As points $1,2$ and $3$ are equidistant from the axis. Thus, $B_1=B_2=B_3$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.