Question

derige the formula of biot-Savart law???,

Answer 1

The Biot-Savart law starts with the following equation: →B=μ04π∫wireId→l׈rr2. B=μ04π∫wireIrdθr2. The current and radius can be pulled out of the integral because they are the same regardless of where we are on the path.

Answer 2

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A current element is a conductor carrying current.It is the product of current,I and length of very small segment of current carrying wire ,dL.

Let us consider a small element AB of length dl of the conductor RS carrying a current I.

Let r be the position vector of the point P from the current element I dL.and θ be the angle dl and r.

According to Biot-Savarts law,the magnetic field induction dB or magnetic flux density at a point P due to current element depends upon the following factors.

(i) dBI

(ii) dBdl

(iii) dBsinθ

(iv) dB1/r^2

Combining these factors,we get

dBIdLsinθ/r^2

or dB=K Idl sinθ/r^2

where K is a constant of perportionality.

In S.I units, K=μ0/4

thus , dB=μ0/4 I dl sinθ/r^2

where μ0 is absolute premeability of free space and

μ0=4*10^-7 Wb A^-1m^-1

= 4*10^-7*TA^-1m [ 1T=1 Wb m^-2]

In C.G.S units,K=1 (In free space)

Thus dB=Idl sinθ/r^2

In vector form,

dB=μ0/4 I(dl*r)/r^3

magnetic field induction at point P due to current through entire wire is

B=∫μ0/4Idl*r/r^3

Or B=∫μ0/4 Idl sin θ/r^2