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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 6 of 35
Marks: +1, -0
A long straight wire of radius ' aa ' carries a steady current II. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic field B→1\overset{\rightarrow_{1}}{B} at   a2\;\frac{a}{2} and B→2\overset{\rightarrow_{2}}{B} at distance 2a2 a is
Magnetic field inside
B1=  μ0ir2πa2=  μ0ia22πa2=  μ0i4πaB_1 = \;\frac{\mu_0 i r}{2 \pi a^2} = \;\frac{\mu_0 i \frac{a}{2}}{2 \pi a^2} = \;\frac{\mu_0 i}{4 \pi a} (Here, r=a2r = \frac{a}{2} )
Magnetic field outside =B2=  μ0i2πr=  μ0i2π(2a)=  μ0i4πa= B_2 = \;\frac{\mu_0 i}{2 \pi r} = \;\frac{\mu_0 i}{2 \pi (2 a)} = \;\frac{\mu_0 i}{4 \pi a} (Here, r=2ar = 2 a )
So, The required ratio   B1B2=1\;\frac{B_1}{B_2} = 1
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