drive the expression for the magnetic field at a distance from a long straight wire carring current by amperes circuital law.
Answer (1)
Hello there!
Greetings!
For deriving the equation , let us first know what is Ampere's Circuital Law, it states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.
Now let us derive the expression.
Let us consider a straight conductor XY carrying current I. We wish to find its magnetic field at the point P whose perpendicular distance from the wire is a i.e. PQ=a
Let a small current element dl of the conductor at O. Its distance from Q is I. i.e OQ=I. Let r be the position vector of the point P relative to the current element and be the angle between dl and r. According to Biot savart law, the magnitude of the field dB due to the current element dl will be
dB=04.Idlsinr2
From right OQP,
+=900
or =900
sin=(900)=cos
Also cos=ar
or r=acos=asec
As tan=la
l=atan
On differentiating, we get
dl=asec2d
Hence dB=04I(asec2d)cosa2sec2
or dB=0I4acosd
According to right hand rule, the direction of the magnetic field at the P due to all such elements will be in the same direction, namely; normally into the plane of paper. Hence the total field B at the point P due to the entire conductor is obtained by integrating the above equation within the limits 1 and 2.
B=21dB=0I4a21cosd
=0I4a[sin]21
=0I4a [sin2sin(1)]
or
B=0I4a [sin2+sin1]
Above equation gives magnetic field due to a finite wire in terms of te angles subtended at the observation point by the ends of wire.
Thankyou
Greetings!
For deriving the equation , let us first know what is Ampere's Circuital Law, it states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.
Now let us derive the expression.
Let us consider a straight conductor XY carrying current I. We wish to find its magnetic field at the point P whose perpendicular distance from the wire is a i.e. PQ=a
Let a small current element dl of the conductor at O. Its distance from Q is I. i.e OQ=I. Let r be the position vector of the point P relative to the current element and be the angle between dl and r. According to Biot savart law, the magnitude of the field dB due to the current element dl will be
dB=04.Idlsinr2
From right OQP,
+=900
or =900
sin=(900)=cos
Also cos=ar
or r=acos=asec
As tan=la
l=atan
On differentiating, we get
dl=asec2d
Hence dB=04I(asec2d)cosa2sec2
or dB=0I4acosd
According to right hand rule, the direction of the magnetic field at the P due to all such elements will be in the same direction, namely; normally into the plane of paper. Hence the total field B at the point P due to the entire conductor is obtained by integrating the above equation within the limits 1 and 2.
B=21dB=0I4a21cosd
=0I4a[sin]21
=0I4a [sin2sin(1)]
or
B=0I4a [sin2+sin1]
Above equation gives magnetic field due to a finite wire in terms of te angles subtended at the observation point by the ends of wire.
Thankyou
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