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how bar magnet is equivalent to solenoid
Asked by munaxa013 | 17 Jan, 2024, 11:56: AM

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 smaller
solenoids with weaker magnetic properties.

The field lines remain continuous, emerging 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 a current-carrying finite solenoid and noting
that the deflections of the needle are similar in both cases.

Magnetic field induction B at a point P on axial line which is at a distance r from centre of a solenoid is

........................... (1)

where μo is permeability of free space , n is number of turns per unit length , is current passing through solenoid,

is length of solenoid and a is radius of solenoid .

Magnetic field induction Bb at a point P on axial line which is at a distance r from centre of a bar magnet is

..................... (2)

If we compare eqn.(1) and (2) , then we find solenoid has magnetic dipole moment m that is given by

m = ( no. of turns ) × Current × ( cross section area ) = n × 2 × × ( πa2 ) ;

Thus, a bar magnet and a solenoid produce similar magnetic fields.

The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid

that produces the same magnetic field.

Answered by Thiyagarajan K | 17 Jan, 2024, 03:43: PM
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