Question

derive an expression for the magnetic field intensity at a point due to current carrying straight wire of infinite length? ​

Answer 1

Answer:

Answer

Ampere's law equation

∮

B

→

.

dℓ

→

=μ∑i

Here

B

→

.

ℓ

→

=μ∑i

here

dμ

→

length element perpendicular to the plane of paper

∮

B

→

.

dℓ

→

=μ

o

∑i

B → magnetic field which is also perpendicular to plane of paper & align with length elements

∮

B

→

.

dℓ

→

=μ

o

∑i

B.∮dℓ=μ

o

i

B.2π=μ

o

i

B=

2πr

μ

o

i

Answer 2

Explanation:

A long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r<a and r>a where a is the distance at which the magnetic field is calculated.

Solution (a) Consider the case r>a. The outer Amperian loop,is a circle concentric with the cross-section. For this loop,

L=2πr

I

e

= Current enclosed by the loop = I

The result is the familiar expression for a long straight wire

B(2πr)=μ

0

I

B=

2πr

μ

0

I

B∝

r

1

(r>a).

(b) Consider the case r<a i.e. the inner Amperian loop .For this loop, taking the radius of the circle to be r, L=2πr

Now the current enclosed I

e

is not I, but is less than this value.Since the current distribution is uniform, the current enclosed is,

I

e

=I(

πa

2

πr

2

)=I

a

2

r

2

Using Amperes law,

B(2rπ)=μ

0

I

a

2

r

2

B=μ

0

I

2πa

2

r

B∝r (r<a)