My book says that an inductor produces magnetic field around it and it stores energy in this field but then i thought a current carrying wire also produces magnetic field around it then why does it not store energy ? And if it does then why do we use inductors ?
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A straight wire carrying a current does indeed store energy in a magnetic field so it does have an inductance. For example see Derivation of self-inductance of a long wire.
However the inductance of a straight wire is very small. Coiling the wire into a solenoid allows you to create a circuit element with a large inductance for a small size.
The inductance of a straight wire is given by:
$$ L_\text{wire} = \frac{\mu\ell}{8\pi} \tag{1}$$
The inductance of a coiled wire is normally written in the form:
$$ L_\text{coil} = \frac{\mu N^2 A}{d} $$
where $N$ is the number of turns in the coil, $A$ is the area of the coil and $d$ is the length of the coil. To compare this with equation (1) we note that $A=C^2/4\pi$ where $C$ is the circumference, and $NC$ is the total length of the wire. Substituting this into the equation above gives:
$$ L_\text{coil} = \frac{\mu \ell^2}{4\pi d} \tag{2} $$
Or we for comparison with equation (1) we could use $\ell=N2\pi r$ to rewrite this as:
$$ L_\text{coil} = \frac{\mu \ell}{2}\frac{Nr}{d} \tag{3} $$
John Rennie
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To clarify, is the inductance of (say) 100m of wire constant, irrespective of whether it is straight or coiled? – Sep 28 '16 at 14:59
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Thank you ! I just want o ask one more question. My book also says that some energy by the battery is spent in working against the induced emf of the inductor and that energy is stored in magnetic field . So if wire also stores energy in magnetic field then against what do we spend energy that is stored in the magnetic field of the wire as in the case of inductors we spend energy against the emf ? Does wire also have an induced emf ? – Varun Chandra Sep 28 '16 at 15:06
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1@VarunChandra: Yes, but the inductance of a straight wire is so small that under normal circumstances it can be ignored. – John Rennie Sep 28 '16 at 15:16
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@JohnRennie: But how can a straight wire have an emf induced in it? , I mean there has to be some changing magnetic flux through the wire for inducing emf and a wire is not a closed curve through which I can associate magnetic flux – Varun Chandra Sep 28 '16 at 15:33
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1@VarunChandra The complete circuit is always a closed curve, otherwise current can't flow around it. If you try to think about "a straight wire carrying current" but ignore how the current gets into one end of the wire and out of the other, you are leaving out an important part of the physics. For a coiled wire inductor you can ignore the rest of the circuit, because the inductance of the connecting wires will (usually) be small compared with the inductor itself. – alephzero Sep 29 '16 at 01:00
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Self-inductance of a straight wire of length $\ell$ is not $\frac{\mu\ell}{8\pi}$, that's just the contribution to self-inductance inferred from magnetic energy that is inside a cylindrical wire with uniform current density on the cross-section. It's a dubious concept, as the contribution due to magnetic energy outside the wire is ignored - this contribution should dominate, as can be seen from the case of infinite straight wire, and the case current is localized on wire surface. Inductance per unit length of a finite wire should depend on $\ell$, and diverge to infinity as $\ell\to \infty$. – Ján Lalinský Feb 25 '23 at 20:01
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For an attempt to calculate self-inductance of a wire of finite length, see (formula 11) https://g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf . The calculation is a little strange due to weighing magnetic flux inside the wire, but (I think correctly) does not forget to take into account magnetic flux outside the wire. – Ján Lalinský Feb 25 '23 at 20:14