An iron rod of volume $10^{-3} \mathrm{~m}^{3}$ and relative permeability 1000 is placed as core in a solenoid with $10 \mathrm{turns} / \mathrm{cm}$. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be:

(D) $5 \times 10^{2} \mathrm{Am}^{2}$

(A) $0.5 \times 10^{2} \mathrm{Am}^{2}$

(B) $50 \times 10^{2} \mathrm{Am}^{2}$

(C) $500 \times 10^{2} \mathrm{Am}^{2}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Magnetic Properties of Matter

An iron rod of volume

1 0^{- 3} m^{3}

and relative permeability 1000 is placed as core in a solenoid with

10 turns / cm

. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be:

(A) $0.5 \times 1 0^{2} Am^{2}$
(B) $50 \times 1 0^{2} Am^{2}$
(C) $500 \times 1 0^{2} Am^{2}$
(D) $5 \times 1 0^{2} Am^{2}$ ✓ Correct

Answer: (D)

Solution

The magnetic moment of the magnetised core is

M = μ_{r} n i V

, where

n

is the number of turns per unit length. The winding density

10

turns/cm is

1000

turns/m. With

μ_{r} = 1000

i = 0.5

A and

V = 1 0^{- 3}

^{3}

M = 1000 \times 1000 \times 0.5 \times 1 0^{- 3} = 500 = 5 \times 1 0^{2}

A m

^{2}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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