How does the magnetic field intensity at the centre of a circular coil carrying current change, if the current through the coil is doubled and the radius of the coil is halved?

Hint: In order to solve this question, we will first know the general expression of magnetic field due to current carrying circular coil at its centre and then by changing the parameters value as given in the question, we will compare both magnetic field and figure out correct value of magnetic field after the changes.

Formula used:
The general magnetic field due to circular current carrying coil of radius ‘R’ and having current of ‘I” at its centre is given by,
$B = \dfrac{{{\mu _o}I}}{{2R}}$
where ${\mu _o} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}$ is known as the permeability of free space.

Complete step by step answer:
According to the question, let us suppose the initial magnetic field without changes at the centre of the coil is $B = \dfrac{{{\mu _o}I}}{{2R}}$ . Now,let the new radius of coil be R’ and the radius is halved of initial value so $R' = \dfrac{R}{2}$. Let new current in the coil be I’ and current is being double of initial value so $I' = 2I$ and now let us say new magnetic field B’ after modification is given by,
$B' = \dfrac{{{\mu _o}I'}}{{2R'}}$
So, on putting the value of parameters we get,
$B' = \dfrac{{{\mu _o}4I}}{{2R}}$
$\Rightarrow B' = \dfrac{{4{\mu _o}I}}{{2R}}$ and initial magnetic field is $B = \dfrac{{{\mu _o}I}}{{2R}}$ so we can write,
$\therefore B' = 4B$

Hence, after modifications magnetic field will be four times to that of initial value.

Note: It should be remembered that, magnetic field at centre of circular coil also depends upon the number of turns in the coil if there are N number of turns in the coil then net magnetic field at centre will be $B = \dfrac{{{\mu _o}NI}}{{2R}}$ and in given question, its assumed that number of turns kept unchanged. The SI unit of magnetic field is known as Tesla denoted by T.