A long insulated copper wire is closely wound as a spiral of 'N' turns. The spiral has inner radius ` a' and outer radius
′
b
′
. The spiral lies in the X−Y plane and a steady current 'I' flows through the wire. The Z-component of the magnetic field at the center of the spiral is :
Answers
Answer:
Explanation:
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
Answer:
Explanation:
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