The electric current in a circular coil of four turns produces a...

Magnetic field at the centre of the coil is given by, $B = \dfrac{{\mu}_{0}i}{2R}$.

Now in the first case:

$B = 4\left(\dfrac{{\mu}_{0}i}{2R}\right)$

and for the second case:

$B ' = \dfrac{{\mu}_{0}i}{2R '}$.

As the length of the wire will remain same, therefore

$2\piR ' = 4\left(2\piR\right) ⇒ R ' = 4R$

Hence, required ratio will be

$\dfrac{B '}{B} = \dfrac{R}{4{R}^{'}} = \dfrac{1}{16}$

$⇒ B ' = 2T$