Question

Match the entries of column I with the entries of column II in the following table.

Column I	Column II
$\left(A\right)$ Magnetic field due to infinite long wire	$\left(P\right)$ $B=\frac{{\mu }_{0}I}{2\pi d}$
$\left(B\right)$ Magnetic field due to current element	$\left(Q\right)$ $B=\frac{{\mu }_{0}IN}{2R}$
$\left(C\right)$ Magnetic field at centre of current carrying coil	$\left(R\right)$ $B=\frac{{\mu }_{0}IN{R}^2}{2(x^2+R^2{)}^{3/2}}$
$\left(D\right)$ Magnetic field at axis of coil	$\left(S\right)$ $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

Which of the following option has the correct combination, considering column I and column II?

• A. $A\to P$
• B. $B\to Q$
• C. $C\to R$
• D. $D\to S$

Answer 1

The correct option is A $A\to P$

We know that, magnetic field due to -

$\left(1\right)$ Infinite long wire : $B=\frac{{\mu }_{0}I}{2\pi d}$

$\therefore A\to P$

$\left(2\right)$ Current element : $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

$\therefore B\to S$

$\left(3\right)$ Current carrying coil at its centre : $B=\frac{{\mu }_{0}IN}{2R}$

$\therefore C\to Q$

$\left(4\right)$ Current carrying coil at its axis : $B=\frac{{\mu }_{0}IN{R}^2}{2(x^2+R^2{)}^{3/2}}$

$\therefore D\to R$

Hence, option $\left(A\right)$ is correct.