How does the magnetic field at a centre of a current-carrying coil change when the radius of the coil is doubled while keeping other parameters constant?

The magnetic field in the centre of a circular coil can be calculated as follows:

$\mathrm{B}=\frac{{\mu }_0\mathrm{I}}{2\mathrm{R}}$

 Where B= the strength of the magnetic field, I= current, and R= radius of the circular coil, ${\mu }_0$= permeability.

The new magnetic field $B^{\text{'}}$ at the centre of a circular coil when its radius doubles will be: 

$B^{\text{'}}=\frac{{\mu }_{0}I}{2R^{\text{'}}}$ now $R^{\text{'}} = 2R$ then, 

$B^{\text{'}}=\frac{{\mu }_{0}I}{2{\displaystyle \left(2R\right)}}= \frac{1}{2}\left(\frac{{\mu }_{0}I}{2{\displaystyle R}}\right)$ 

$B^{\text{'}}=\frac{B}{2}$

The magnetic field and radius of the coil are inversely related. 

Therefore, when the radius is doubled, the magnetic field at the centre will become half.