JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Magnetic Properties of Matter

A long solenoid with

1000 turns / m

has a core material with relative permeability 500 and volume

1 0^{3} cm^{3}

. If the core material is replaced by another material having relative permeability of 750 with same volume maintaining same current of 0.75 A in the solenoid, the fractional change in the magnetic moment of the core would be approximately

(\frac{x}{499})

. Find the value of

x

Answer: 250

Solution

The magnetic moment of the core is

m = M V

, where

M

is the magnetization and

V

is the volume. For a solenoid,

M = χH = (μ_{r} - 1) H

, where

H = n I

is the magnetic field intensity.

Initially,

m_{1} = (μ_{r 1} - 1) n I V = (500 - 1) \times 1000 \times 0.75 \times 1 0^{- 3} = 499 \times 750 \times 1 0^{- 3}

.

After replacement,

m_{2} = (μ_{r 2} - 1) n I V = (750 - 1) \times 1000 \times 0.75 \times 1 0^{- 3} = 749 \times 750 \times 1 0^{- 3}

.

Fractional change

= \frac{m _{2} - m _{1}}{m _{1}} = \frac{749 - 499}{499} = \frac{250}{499}

.

Thus,

x = 250

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Magnetic Properties of Matter

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