Calculate the magnetic field generated at a distance x from the center of a finite straight wire with a steady current I, and length "2a". Take the limit as the length of the wire goes to infinity and compare this result with that obtained using Ampere's law



Answer :

Answer:To calculate the magnetic field generated at a distance

x from the center of a finite straight wire with a steady current

I and length

2

2a, we can use the Biot-Savart law. Let's consider the wire to be along the

z-axis, with the center of the wire at the origin, and

−a≤z≤a.

Explanation:Biot-Savart Law

The Biot-Savart law states that the magnetic field

B at a point due to a current element

Idl is given by:

=

0

4

×

3

dB=

μ

0

I

 

∣r∣

3

dl×r

where

r is the vector from the current element to the point where the magnetic field is being calculated.

For a straight wire, consider a current element

Idz at position

z on the wire. The position vector from this element to the point where we are calculating the magnetic field (at distance

x from the wire) is:

=

^

^

r=x

x

^

−z

z

^

The differential length element is:

=

^

dl=dz

z

^

The cross product

×

dl×r is:

×

=

^

×

(

^

^

)

=

^

dl×r=dz

z

^

×(x

x

^

−z

z

^

)=xdz

y

^

The magnitude of

r is:

=

2

+

2

∣r∣=

x

2

+z

2

Thus, the differential magnetic field is:

=

0

4

^

(

2

+

2

)

3

/

2

dB=

μ

0

I

 

(x

2

+z

2

)

3/2

xdz

y

^

Integrate this from

=

z=−a to

=

z=a to find the total magnetic field at distance

x from the wire:

=

0

4

(

2

+

2

)

3

/

2

B=

μ

0

Ix

−a

a

 

(x

2

+z

2

)

3/2

dz

Solving the Integral

Let's solve the integral:

(

2

+

2

)

3

/

2

−a

a

 

(x

2

+z

2

)

3/2

dz

This can be done using the substitution

=

tan

(

)

z=xtan(θ):

=

sec

2

(

)

dz=xsec

2

(θ)dθ

2

+

2

=

2

sec

2

(

)

x

2

+z

2

=x

2

sec

2

(θ)

(

2

+

2

)

3

/

2

=

3

sec

3

(

)

(x

2

+z

2

)

3/2

=x

3

sec

3

(θ)

The integral limits change accordingly. When

=

z=−a,

=

tan

1

(

/

)

θ=−tan

−1

(a/x). When

=

z=a,

=

tan

1

(

/

)

θ=tan

−1

(a/x). So,

(

2

+

2

)

3

/

2

=

tan

1

(

/

)

tan

1

(

/

)

sec

2

(

)

3

sec

3

(

)

=

1

2

tan

1

(

/

)

tan

1

(

/

)

cos

(

)

−a

a

 

(x

2

+z

2

)

3/2

dz

=∫

−tan

−1

(a/x)

tan

−1

(a/x)

 

x

3

sec

3

(θ)

xsec

2

(θ)dθ

=

x

2

1

−tan

−1

(a/x)

tan

−1

(a/x)

cos(θ)dθ

=

1

2

[

sin

(

)

]

tan

1

(

/

)

tan

1

(

/

)

=

1

2

(

sin

(

tan

1

(

)

)

sin

(

tan

1

(

)

)

)

=

x

2

1

[sin(θ)]

−tan

−1

(a/x)

tan

−1

(a/x)

=

x

2

1

(sin(tan

−1

(

x

a

))−sin(−tan

−1

(

x

a

)))

Since

sin

(

tan

1

(

/

)

)

=

2

+

2

sin(tan

−1

(a/x))=

x

2

+a

2

a

, we get:

(

2

+

2

)

3

/

2

=

2

2

2

+

2

−a

a

 

(x

2

+z

2

)

3/2

dz

=

x

2

 

x

2

+a

2

2a

Magnetic Field

Thus, the magnetic field at a distance

x from the center of the wire is:

=

0

4

2

2

2

+

2

=

0

2

(

2

+

2

)

1

/

2

B=

μ

0

Ix

x

2

 

x

2

+a

2

2a

=

2πx(x

2

+a

2

)

1/2

μ

0

Ia

Limit as Length Goes to Infinity

Take the limit as

a→∞:

=

lim

0

2

(

2

+

2

)

1

/

2

B=lim

a→∞

 

2πx(x

2

+a

2

)

1/2

μ

0

Ia

For large

a:

(

2

+

2

)

1

/

2

(x

2

+a

2

)

1/2

≈a

So,

=

lim

0

2

=

0

2

B=lim

a→∞

 

2πxa

μ

0

Ia

=

2πx

μ

0

I

Comparison with Ampere's Law

Using Ampere's Law for an infinite straight wire, the magnetic field at a distance

x from the wire is:

=

0

2

B=

2πx

μ

0

I

This result matches the one obtained by taking the limit of the finite wire's magnetic field as its length goes to infinity. Thus, the calculation using the Biot-Savart law and the result from Ampere's law are consistent.