MHT CET · Physics · Magnetic Properties of Matter

The magnetization of bar magnet of length 5 cm, cross sectional area 2 $c m^{2}$ and net magnetic moment $1 A m^{2}$ is

A $3 \times \frac{10^{5} A}{m}$
B $4 \times \frac{10^{5} A}{m}$
C $2 \times \frac{10^{5} A}{m}$
D $1 \times \frac{10^{5} A}{m}$

Verified Solution

Answer & Solution

Correct Answer

(D) $1 \times \frac{10^{5} A}{m}$

Step-by-step Solution

Detailed explanation

Given,

l = 5 c m = 5 \times 10^{- 2} m,

a = 2 c m^{2} = 2 \times 10^{- 4} m^{2}

And

M = 1 A m^{- 2}

The magnetisation of the bar magnet is given by

l = \frac{M}{V}

where, V = volume of bar magnet =

a \times l

\Rightarrow l = \frac{M}{a \times l}

Substituting given values, we get

l = \frac{1}{2 \times 10^{- 4} \times 5 \times 10^{- 2}} = \frac{1}{10^{- 5}}

= 1 \times \frac{10^{5} A}{m}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The magnetization of bar magnet of length 5 cm, cross sectional area 2 cm2 and net magnetic moment 1 Am2 is

Step-by-step Solution

The magnetization of bar magnet of length 5 cm, cross sectional area 2 $c m^{2}$ and net magnetic moment $1 A m^{2}$ is