5.2.2 Bar magnet as an equivalent solenoid
In the previous chapter, we have explained how a current loop acts assz magnetic dipole (Section 4.10). We mentioned Ampere's hypo all magnetic phenomena can be explained in terms of circulationgestic moment in associated with a current loop was deffined to be m=NI A where N is the number of turns in the loop, I the current and A the area vector (Eq. 4.30 ).

The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smallet solenoids with weaker magnetic propertiess The field lines remain continuous, emerglifg
(a) from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighbourhood of a bar magnet and current-carrying finite solenoid and notifts that the deflections of the needle are similis in both cases.
(b) To make this analogy more firm we 5.4 Calculation of (a) The axial field of a calculate the axial field of a finite solenoli) plenoid in order to demonstrate its similarity depicted in Fig. 5.4 (a). We shall demonstrally at of a bar magnet. (b) A magnetic needle that at large distances this axial fill in a uniform magnetic field B. The resembles that of a bar magnet. arrangement may be used to
Let the solenoid of Fig. 5.4(a) consists? determine either B or the magnetic n turns per unit length. Let its length be? moment m of the needle. and radius a. We can evaluate the axal fic at a point P, at a distance r from the centri of the solenoid. To do this, consider a circular element of thickness dre the solenold at a distance x from its centre. It consists of ndx turns. it be the current in the solenoid. In Section 4.6 of the previous chapter have calculated the magnetic field on the axis of a circular current loo From Eq. (4.13), the magnitude of the field at point P due to the circult element is

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5.2.2 Bar magnet as an equivalent solenoid
In the previous chapter, we have explained how a current loop acts assz magnetic dipole (Section 4.10). We mentioned Ampere's hypo all magnetic phenomena can be explained in terms of circulationgestic moment in associated with a current loop was deffined to be m=NI A where N is the number of turns in the loop, I the current and A the area vector (Eq. 4.30 ).

The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smallet solenoids with weaker magnetic propertiess The field lines remain continuous, emerglifg
(a) from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighbourhood of a bar magnet and current-carrying finite solenoid and notifts that the deflections of the needle are similis in both cases.
(b) To make this analogy more firm we 5.4 Calculation of (a) The axial field of a calculate the axial field of a finite solenoli) plenoid in order to demonstrate its similarity depicted in Fig. 5.4 (a). We shall demonstrally at of a bar magnet. (b) A magnetic needle that at large distances this axial fill in a uniform magnetic field B. The resembles that of a bar magnet. arrangement may be used to
Let the solenoid of Fig. 5.4(a) consists? determine either B or the magnetic n turns per unit length. Let its length be? moment m of the needle.  and radius a. We can evaluate the axal fic at a point P, at a distance r from the centri of the solenoid. To do this, consider a circular element of thickness dre the solenold at a distance x from its centre. It consists of ndx turns. it be the current in the solenoid. In Section 4.6 of the previous chapter have calculated the magnetic field on the axis of a circular current loo From Eq. (4.13), the magnitude of the field at point P due to the circult element is

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Question Text	5.2.2 Bar magnet as an equivalent solenoid In the previous chapter, we have explained how a current loop acts assz magnetic dipole (Section 4.10). We mentioned Ampere's hypo all magnetic phenomena can be explained in terms of circulationgestic moment in associated with a current loop was deffined to be $m = N I$ A where $N$ is the number of turns in the loop, $I$ the current and $A$ the area vector (Eq. $4.30$ ). The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smallet solenoids with weaker magnetic propertiess The field lines remain continuous, emerglifg (a) from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighbourhood of a bar magnet and current-carrying finite solenoid and notifts that the deflections of the needle are similis in both cases. (b) To make this analogy more firm we 5.4 Calculation of (a) The axial field of a calculate the axial field of a finite solenoli) plenoid in order to demonstrate its similarity depicted in Fig. $5.4$ (a). We shall demonstrally at of a bar magnet. (b) A magnetic needle that at large distances this axial fill in a uniform magnetic field $B$ . The resembles that of a bar magnet. arrangement may be used to Let the solenoid of Fig. 5.4(a) consists? determine either $B$ or the magnetic $n$ turns per unit length. Let its length be? moment $m$ of the needle. and radius a. We can evaluate the axal fic at a point $P$ , at a distance $r$ from the centri of the solenoid. To do this, consider a circular element of thickness dre the solenold at a distance $x$ from its centre. It consists of $n d x$ turns. it be the current in the solenoid. In Section $4.6$ of the previous chapter have calculated the magnetic field on the axis of a circular current loo From Eq. (4.13), the magnitude of the field at point $P$ due to the circult element is
Updated On	Jan 6, 2023
Topic	Magnetism, EMI and AC
Subject	Physics
Class	Class 12
Answer Type	Video solution: 1
Upvotes	124
Avg. Video Duration	8 min

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