A circular coil carrying a current I has radius R and number of turns N . If all the three, i.e., the current I, radius R and number of turns N are doubled, then magnetic field at its centre becomes

 When current I flows in a circular coil with radius R and number of turns N, we first apply the Biot-Savart law to calculate the relationship between magnetic field, radius, current, and number of turns, i.e.

We know that at the axis of a current carrying coil at distance x from the centre, magnetic field B is given by

B=μ0NIR2/2(R2+x2)3/2

At centre x=0 So,

BC=μ0NIR2/2(R2)3/2

BC=μ0NIR2/2R3

⟹ BC = μ0IN/2R …..(1)

If I, R & N are doubled then magnetic field at centre will be BC′ = μ0(2I)(2N)/2(2R)

⟹BC′ = 2(μ0NI/2R)

From equation (1) BC′ = 2BC

Hence, the magnetic field at the centre will also become doubled.