Question

A current $I$ flows along a thin walled tube of radius $R$ with a long longitudinal slit of width $b(<<R)$. What is the value of magnetic field at the centre.

• A. $\frac{{\mu }_{0}Ib}{4{\pi }^2{R}^2}$
• B. $\frac{{\mu }_{0}Ib}{2{\pi }^2{R}^2}$
• C. Zero
• D. $\frac{{\mu }_{0}I}{4{\pi }^2{R}^2}$

Answer 1

The correct option is A $\frac{{\mu }_{0}Ib}{4{\pi }^2{R}^2}$

The point lying at the centre is inside the thin walled tube.

If the tube would have been complete, then from Ampere's circuital law

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{en}$

$\because I_{en}=0$

$\therefore B_{net}=0$ at the centre.

The current is flowing along wall of the tube, hence current flowing per unit length on circumference is $\frac{I}{(2\pi R-b)}=\frac{I}{2\pi R}(\because R>>b)$

Now the amount of current flowing through an imaginary strip of width $b$ in order to complete the whole tube will be

$I^{\prime }=\frac{I}{\left(2\pi R\right)}\times b=\frac{Ib}{\left(2\pi R\right)}$

Now using superposition principle,

$B_1+B_2=0$

$(B_1$ = magnetic field due to strip of width b)

$B_2$ = magnetic field due to given tube

$\Rightarrow B_2=-B_1$

Now $B_1=\frac{{\mu }_0{I}^{\prime }}{2\pi R}$ ( $\because R$ is distance of centre from slit)

$\Rightarrow B_1=\frac{{\mu }_{0}Ib}{2\pi R\left(2\pi R\right)}$

Hence, $B_2=-B_1$

$\therefore |B_2|=|B_1|=\frac{{\mu }_{0}Ib}{4{\pi }^2{R}^2}$