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A 25cm long solenoid has radius 2cm and 500 total number of turns. It carries a current of 15A. If it is equivalent to magnet of the same size and magnetization $ \overrightarrow{\text{M}}\left( \dfrac{\text{Magnetic moment}}{\text{volume}} \right) $ , then $ \overrightarrow{\left| \text{M} \right|} $ is
(A) $ \text{3}00\text{ A}{{\text{m}}^{-\text{1}}} $
(B) $ \text{3}0,000\text{ A}{{\text{m}}^{-\text{1}}} $
(C) $ \text{3}0,000\text{A}{{\text{m}}^{-\text{1}}} $
(D) $ \text{3A}{{\text{m}}^{-\text{1}}} $

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Last updated date: 25th Feb 2024
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Answer
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Hint: We use here magnetization formula which is equal to ratio of magnetic moment to volume. The magnetic moment of solenoid is given by NIA where
N is total number of turns,
I is current in coil,
A is an area of the coil of solenoid.
Put all the values given in question in the above formula, And we will get the required result.

Complete step by step solution
Magnetization can be defined as the net magnetic moment per unit volume of material.
 $ \text{M}=\dfrac{\text{magnetic moment}}{\text{volume of material}} $ -------- (1)
As the solenoid is equivalent of magnet of same size, therefore magnetic moment of solenoid is given by = NIA
Here, N is no. of turns =500
I is current of loop = 15A
A is the area of the coil of solenoid.
Since, volume can be written as
= Area $ \times $ Length
V = AL
L is the length of solenoid = 25 cm
 $ =\text{ 25}\times \text{1}{{0}^{-\text{2}}}\text{m} $
 $ =\text{ }0\cdot \text{25m} $
Use the above value in equation (1)
 $ \text{M=}\dfrac{\text{NIA}}{\text{AL}} $
 $ \text{M=}\dfrac{\text{NI}}{\text{L}} $
  $ \text{M}=\dfrac{500\times 15}{0.25} $ $ \text{M}=\dfrac{500\times 15\times 100}{25} $
 $ \text{M=30,000A}{{\text{m}}^{\text{-1}}} $
This is the value of magnetization.

Note
A linear solenoid carrying current is equivalent to a bar magnet.
The solenoid field lines due to current carrying solenoid resemble exactly with those of bar magnet.
The magnetic field induction at a point near the middle just outside the curved face of the solenoid carrying current is zero.
The net force of the magnetic dipole inside the solenoid carrying current is zero.
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