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.
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Physics