JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Magnetic Properties of Matter

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A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If

B_{H} = 0.4 G

, the magnetic moment of the magnet is (

1 G = 1 0^{- 4} T

)

(A) $2.880 \times 1 0^{3} JT^{- 1}$
(B) $2.880 JT^{- 1}$ ✓ Correct
(C) $2.880 \times 1 0^{2} JT^{- 1}$
(D) $28.80 JT^{- 1}$

Answer: (B)

Solution

With the north pole of the magnet pointing geographic north, the magnet's axial field points north as well and adds to the earth's horizontal field, so no neutral point can lie on the axis. On the equatorial line the magnet's field points south, and a neutral point forms where it exactly cancels

B_{H}

.
For a bar magnet of half length

l = 7

cm, at an equatorial distance

r = 18

cm,

\frac{μ _{0}}{4 π} \cdot \frac{M}{( r ^{2} + l ^{2} ) ^{3/2}} = B_{H}

With

\frac{μ _{0}}{4 π} = 1 0^{- 7}

T m A

^{- 1}

and

B_{H} = 0.4 G = 4 \times 1 0^{- 5}

M = \frac{B _{H} ( r ^{2} + l ^{2} ) ^{3/2}}{1 0 ^{- 7}} = \frac{4 \times 1 0 ^{- 5} \times ( 0.0324 + 0.0049 ) ^{3/2}}{1 0 ^{- 7}} = \frac{4 \times 1 0 ^{- 5} \times 7.204 \times 1 0 ^{- 3}}{1 0 ^{- 7}}

The magnetic moment of the magnet is

2.880 J T^{- 1}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Magnetic Properties of Matter

Solution

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