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Bar Magnet As An Equivalent Solenoid

Bar Magnet As An Equivalent Solenoid :- The similarity between the magnetic field lines of a bar magnet and the magnetic field lines of a current carrying solenoid suggests that just as a solenoid is the sum of many circulating currents, similarly a bar magnet can also be the sum of many circulating currents.

Just as by cutting a bar magnet into two equal pieces, we get two small bar magnets with relatively weak magnetic fields, in the same way, by cutting a solenoid, we get two small solenoids.

Bar Magnet As An Equivalent Solenoid – Mathematical Proof

To mathematically prove that a bar magnet behaves like a current carrying solenoid, we calculate the axial magnetic field of a finite length solenoid shown in the following figure. Here we will show that over very large distances the axial magnetic field of the solenoid is the same as the axial magnetic field of a bar magnet.

Suppose in the above figure, the number of turns per unit length of the current carrying solenoid = n, radius = a, length = 2l and electric current flowing in the solenoid = I. We find the axial magnetic field at a point P located at a distance r from the center O of the solenoid. To do this, a small circular part of the solenoid of thickness dx is taken which is located at a distance x from its center O. The total number of turns in this fraction will be ndx.

We know that the magnetic field on the axis of a circular current carrying loop is given by $\displaystyle B=\frac{{{\mu }_{0}}NI{{a}^{2}}}{2{{[{{a}^{2}}+{{x}^{2}}]}^{3/2}}}$ . Here small magnetic field at the observation point P due to a fraction of the thickness dx is given by,

$\displaystyle dB=\frac{{{\mu }_{0}}ndxI{{a}^{2}}}{2{{[{{a}^{2}}+{{(r-x)}^{2}}]}^{3/2}}}$ …..(1)

To obtain the magnitude of the total magnetic field at the observation point P due to the entire solenoid, the contributions of all such minor parts have to be added. In other words, equation (1) has to be integrated from x = – l to x = + l. Therefore

$\displaystyle B=\int\limits_{-l}^{+l}{dB}=\int\limits_{-l}^{+l}{\frac{{{\mu }_{0}}ndxI{{a}^{2}}}{2{{[{{a}^{2}}+{{(r-x)}^{2}}]}^{3/2}}}=}\frac{{{\mu }_{0}}nI{{a}^{2}}}{2}\int\limits_{-l}^{+l}{\frac{dx}{{{[{{a}^{2}}+{{(r-x)}^{2}}]}^{3/2}}}}$ …..(2)

If the observation point P is located at a large distance on the axis of the solenoid, then r >> a and r >> l, hence the value of the denominator in equation (2), will be approximately,

$\displaystyle {{[{{a}^{2}}+{{(r-x)}^{2}}]}^{3/2}}\cong {{r}^{3}}$

Hence from equation (2) we get,

$\displaystyle B=\frac{{{\mu }_{0}}nI{{a}^{2}}}{2{{r}^{3}}}\int\limits_{-l}^{+l}{dx}$

$\displaystyle \Rightarrow B=\frac{{{\mu }_{0}}nI}{2}\frac{2l{{a}^{2}}}{{{r}^{3}}}$ …..(3)

The magnitude of the magnetic moment of a current carrying solenoid,

M = total number of turns × electric current × area of cross section

⇒ M = n × (2l) × I × (πa²)

Hence from equation (3),

$\displaystyle B=\frac{{{\mu }_{0}}}{2\pi }\frac{(n2l)(I)(\pi {{a}^{2}})}{{{r}^{3}}}=\frac{{{\mu }_{0}}}{2\pi }\frac{M}{{{r}^{3}}}$

$\displaystyle \Rightarrow B=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{r}^{3}}}$ …..(4)

If the magnetic field is determined experimentally for a point located far away from the axis of a bar magnet, then an expression similar to equation (4) will be obtained.

Thus, in this article we proved the behavior of a bar magnet like a current carrying solenoid.

Bar Magnet As An Equivalent Solenoid

Bar Magnet As An Equivalent Solenoid – Mathematical Proof

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