AIEEE 2007 Solved Paper

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the ampere-an path formed at a distance r1(=‌

a

2

)
=(‌

πr12

πa2

)×I, where I is total current
∴ Magnetic field at P1 is
B1=‌

µ0×‌ current enclosed ‌

‌ path ‌

⇒B1=‌

µ0×( πr12 πa2 )×I

2πr1

=‌

µ0×Ir1

2πa2

Now, magnetic field at point P2,B2=‌

µ0

2π

⋅‌

I

(2a)

=‌

µ0I

4πa

∴ Required ratio =‌

B1

B2

=‌

µ0Ir1

2πa2

×‌

4πa

µ0I

=‌

2r1

a

=‌

2× a 2

a

=1.