Question

A long straight wire of radius x carries a steady current I. The current is uniformly distributed across its cross section. Calculate the ratio of the magnetic field at x⁴ and 8x.​

Answer 1

Answer:

Solution :

We know,magnetic field B in inside points =

2πR

2

μ

0

ir

[where,R = radius,r = distance of the inside point from the axis and i = total current]

Distance from the axis at

4

a

inside = (a−

4

a

) =

4

3a

So,Magnetic field at

4

a

inside =

2πa

2

μ

0

×i×

4

3a

=

8πa

3μ

0

i

Distance from the axis at

4

a

outside = (a+

4

a

)

We also know,magnetic field B at outside points =

2πr

μ

0

i

[r=distance of the outside point from the axis]

So,magnetic field at

4

a

outside =

2π×

4

5a

μ

0

i

=

5πa

2μ

0

i

The ratio of the magnetic field at

4

a

inside and

4

a

outside the surface of the wire =

5πa

2μ

0

i

8πa

3μ

0

i

=15:16

Answer 2

Given: A long straight wire of radius x carries a steady current I. The current is uniformly distributed across its cross section.

To find: The ratio of the magnetic field at x⁴ and 8x.​

Solution:

• The ampere circuital law states that the linear integral of the field around any closed path is equal to μ₀ times the net current threading through the area enclosed by the path.
• When this law is applied to a long straight wire, the field is given by the formula,

        $B = \frac{\mu _{0} I}{2\pi r}$

• Here, B is the field and r is the radius of the wire.
• Since the change in the field is inversely proportional to the radius, the ratio can be found as follows.

        $\frac{B_{1}}{B_{2}} = \frac{r_{2}}{r_{1}}$

             $= \frac{8x}{x^{4}}$

             $= \frac{8}{x^{3}}$

             $= 8:{x} ^{3}$

Therefore, the ratio of the magnetic field at x⁴ and 8x is 8 : x³.