Derive an expression for the axial magnetic field of a finite solenoid of length 2l and radius r carrying current I. Under what condition does the field become equivalent to that produced by a bar magnet?

Q: Derive an expression for the axial magnetic field of a finite solenoid of length 2l and radius r carrying current I. Under what condition does the field become equivalent to that produced by a bar magnet?

1. The magnitude of the total field is obtained due to small elements $\text{dB} = \frac{\mu_{o}\text{ndxla}^{2}}{2[(\text{r} - \text{x})^{2} + \text{a}^{2}]\frac{3}{2}}$ x varies from $\text{x} = - l \text{ to } \text{x} = + l $ $\text{B} = \frac{\mu_{o}\text{nla}^{2}}{2} \int\limits^{l}_{-l}\frac{\text{dx}}{[(\text{r} - \text{x})^{2} +\text{a}^{2}] \frac{3}{2}}$ For $\text{r} >> \text{a} \text{ and } , \text{we have } \text{r} > > \text{x}$ $\text{B}\simeq\frac{\mu_{o}n\text{la}^{2}}{2\text{r}^3}\int\limits^{l}_{-l}\text{dx} =\text{B} = \frac{\mu_{o}\text{nla}^{2}(2\ell)}{2\text{r}^{3}}$ Here magnetic moment m = n2l $(\pi\text{a}^{2}) $ $\text{Thus }\text{B} = \frac{\mu_{o}2\text{m}}{4\pi\text{r}^{3}}$ This is also the far axial magnetic field of a bar magnet. Hence, the magnetic field, due to current carrying solenoid along its axial line is similar to that of a bar magnet for far off axial points.

Question

CBSE OUTSIDE DELHI - SET 2 SOUTH 2016

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Solution

The magnitude of the total field is obtained due to small elements

$\text{dB} = \frac{\mu_{o}\text{ndxla}^{2}}{2[(\text{r} - \text{x})^{2} + \text{a}^{2}]\frac{3}{2}}$

x varies from $\text{x} = - l \text{ to } \text{x} = + l $

$\text{B} = \frac{\mu_{o}\text{nla}^{2}}{2} \int\limits^{l}_{-l}\frac{\text{dx}}{[(\text{r} - \text{x})^{2} +\text{a}^{2}] \frac{3}{2}}$

For $\text{r} >> \text{a} \text{ and } , \text{we have } \text{r} > > \text{x}$

$\text{B}\simeq\frac{\mu_{o}n\text{la}^{2}}{2\text{r}^3}\int\limits^{l}_{-l}\text{dx} =\text{B} = \frac{\mu_{o}\text{nla}^{2}(2\ell)}{2\text{r}^{3}}$

Here magnetic moment m = n2l $(\pi\text{a}^{2}) $

$\text{Thus }\text{B} = \frac{\mu_{o}2\text{m}}{4\pi\text{r}^{3}}$

This is also the far axial magnetic field of a bar magnet. Hence, the magnetic field, due to current carrying solenoid along its axial line is similar to that of a bar magnet for far off axial points.

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