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Question
Obtain an expression for magnetic flux density B at the centre of a circular coil of radius R, having N turns and carrying a current I
Solution
Magnetic flux density B at the centre of a circular coil of radius r having N turns and carrying a current I.
Consider a circular coil of radius r and carrying the current I in the direction as shown in the figure. Suppose the entire circular coil is divided into a large number of current elements, each of length dl. According to Biot-savart law, the magnetic field `vec(dB)` at the centre O of the coil due to current element `Ivec(dl)` is given by
`vec(dB) = mu_0/(4pi) (I(vec(dl) xx vecr))/r^3`
The magnitude of `vec(db)` at the centre O is
`dB = mu_0/(4pi) (Idl xx r sin theta)/(r^3)`
`db = mu_0/(4pi) (I dl sin theta)/r^2`
`.B = int dB`
`= int mu_0/(4pi) (i dl sin theta)/r^2`
`theta= 90^@` `∴ sin 90^@ = 1`
`:. B = mu_o/(4pi) I/r^2 int dldl`
`intdl`= total length of the coil = `2pir`
`B = mu_o/(4pi) I/r^2 (2pir)`
`B =(mu_oI)/(2r)`
If the coil N turns
`:. B = (mu_oNI)/(2r)`
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