CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

Question : 6 of 35

Marks: +1, -0

A long straight wire of radius '

a

' carries a steady current

I

. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic field

{B}↖{→}_1

\;{a}/{2}

and

{B}↖{→}_2

at distance

2 a

Solution:

Magnetic field inside

B_1=\;{µ_0 i r}/{2 π a^2}=\;{µ_0 i(a \/ 2)}/{2 π a^2}=\;{µ_0 i}/{4 π a}

(Here,

r=a \/ 2

)
Magnetic field outside

=B_2=\;{µ_0 i}/{2 π r}=\;{µ_0 i}/{2 π(2 a)}=\;{µ_0 i}/{4 π a}

(Here,

r=2 a

)
So, The required ratio

\;{B_1}/{B_2}=1

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