A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross-section. Find the ratio of magnetic field at a/2 and 2a.

Solution

Given a long straight wire of radius 'r' carries uniform current.

The current enclosed by first amperian path = $\frac{{Iπr}^{2}_{1}}{{πR}^{2}} = \frac{I {r^{2}}_{1}}{R^{2}}$

Now, using Ampere circuital law,'
$\oint B . d l = μ_{o} i$
$\Rightarrow$ B = $\frac{μ_{o} i}{2 π r_{1}} = \frac{μ_{o}}{2 π r_{1}} \times \frac{I {r_{1}}^{2}}{R^{2}} = \frac{μ_{o} I r_{1}}{2 π R^{2}}$

Magnetic field induction at a distance r₂ is,
B' = $\frac{μ_{o}}{4 π} \frac{2 I}{r_{2}}$

Therefore,
Ratio of magnetic field at a/2 and 2a is,
$\frac{B}{B'} = \frac{r_{1} r_{2}}{R^{2}} = \frac{(\frac{a}{2}) . 2 a}{a^{2}} = 1$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross-section. Find the ratio of magnetic field at a/2 and 2a.

Some More Questions From Moving Charges And Magnetism Chapter

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