a long insulated copper wire is closely wounded as a spiral of n turns. the spiral has inner radius a and outter radius b . the spiral lies in xy plane and steady current I flows through the wire. z component of the magnetic field at the centre of the spiral is
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Answer:
Let us consider an elementary ring of radius r and thickness dr in which current I is flowing.
Number of turns in this elementary ring dN=
b−a
N
dr
Thus magnetic field at the centre O due to this ring dB=
2r
μ
o
IdN
We get dB=
2(b−a)r
μ
o
INdr
Net magnetic field at centre of spiral B=∫
a
b
2(b−a)
μ
o
IN
r
dr
∴ B=
2(b−a)
μ
o
IN
∫
a
b
r
dr
Or B=
2(b−a)
μ
o
IN
×lnr
∣
∣
∣
∣
∣
a
b
Or B=
2(b−a)
μ
o
IN
ln
a
b
solution
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