Get Answers to all your Questions

header-bg qa

A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of magnetic field B and B^{\prime} at radial distance a/2 and 2a respectively, from the axis of the wire is 

Option: 1

1/2


Option: 2

1/4


Option: 3

1/8


Option: 4

1


Answers (1)

best_answer

According to given data B \times 2 \pi\left(\frac{a}{2}\right)=\frac{\mu_0 I}{\pi a^2} \times \pi\left(\frac{a}{2}\right)^2

\Rightarrow \quad B=\frac{\mu_0 I}{4 \pi a}

Similarly, the magnetic field B^{\prime} at radial distance 2a is given by

B^{\prime} \times 2 \pi \times 2 a=\mu_0 I

\Rightarrow B^{\prime}=\frac{\mu_0 I}{4 \pi a}

So \frac{B}{B^{\prime}}=\frac{\mu_0 I / 4 \pi a}{\mu_0 I / 4 a \pi}=1

Posted by

manish

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks