Acircular loop is given in figure with gurrent I and radius R find the value of magnetic induction as its centre
Answers
Answer:
We have a ring of radius R carrying current I as shown in the following figure.
We will take an element on ring of length d
l
,
r
is the position vector of point P from the element d
l
. In the front view, we can see that horizontal component will cancel each other. Only vertical will add up and will give net magnetic field.
Using Biot-savart law:-
Magnetic field at point P due to element d
l
=d
B
=(
4π
μ
0
I
)
r
2
dl
Vertical component =(dB)sinα
Net magnetic field =B
Net
=∫(dB)sinα
⇒B
Net
=∫(
4π
μ
0
I
)
r
2
dl
sinα=
4π
μ
0
I
∫
r
R
×
r
2
1
dl
⇒B
Net
=
4πr
3
μ
0
IR
∫dl;
since ∫dl=2πR
⇒B
Net
=
4πr
3
μ
0
TR
×2πR=
2r
3
μ
0
IR
2
Along axis
Since, ∣
r
∣=r=
R
2
+x
2
∴B
Net
=
2(
R
2
+x
2
)
3
μ
0
IR
2
=(
2(R
2
+x
2
)
3/2
μ
0
R
2
)
Magnetic field at a point on its axis at a distance x from its centre.