The ratio of magnetic fields due to a bar magnet at the two axial points P 1 & and P 2 & which are separated from each other by 10 cm is 25: 2 . P 1~ Point is situated at 10 cm from the centre of the magnet. (Points P 1

Correct answer: B. 10 cm B _ 1 B _ 2 = (25)/(2) μ _ 0 4π M d _ 1 ( d _ 1 ^ 2 - l^ 2 ) ^ 2 μ _ 0 4π M d _ 2 ( d _ 2 ^ 2 - l^ 2 ) ^ 2 = (25)/(2) d _ 1 d _ 2 × ( d _ 2 ^ 2 - l^ 2 ) ^ 2 ( d _ 1 ^ 2 - l^ 2 ) ^ 2 = (25)/(2) d _ 1 = 10cm, d _ 2 = 20cm (10)/(20) × ( 20 ^ 2 - l^ 2 10 ^ 2 - l^ 2 ) ^ 2 = (25)/(2) 400 - l^ 2 = 5( 100 - l^ 2 ) 4l^ 2 = 100 ⇒ l^ 2 = 25 l = 5cm &

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Magnetic Properties of Matter

JEE Main · Physics · Magnetic Properties of Matter

The ratio of magnetic fields due to a bar magnet at the two axial...

The ratio of magnetic fields due to a bar magnet at the two axial points P₁& and P₂& which are separated from each other by

10

cm is

25: 2

. P_1~Point is situated at

10

cm from the centre of the magnet. (Points P₁& and P_2&are on the same side of magnet and distance of P₁ and P₂from the centre of magnet is much greater than the distance of two ends of the magnet from the centre). Magnetic length of the bar magnet is

Options

A $5$ cm
B $10$ cm
C $15$ cm
D $20$ cm

Correct answer

B. $10$ cm

Step-by-step solution

\dfrac{{B}_{1}}{{B}_{2}} = \dfrac{25}{2}

\dfrac{\dfrac{{μ}_{0}}{4\pi} \dfrac{M{d}_{1}}{{( {d}_{1}^{2} - l^{2} )}^{2}}}{\dfrac{{μ}_{0}}{4\pi} \dfrac{M{d}_{2}}{{( {d}_{2}^{2} - l^{2} )}^{2}}} = \dfrac{25}{2}

\dfrac{{d}_{1}}{{d}_{2}} \times \dfrac{{( {d}_{2}^{2} - l^{2} )}^{2}}{{( {d}_{1}^{2} - l^{2} )}^{2}} = \dfrac{25}{2}

{d}_{1} = 10cm, {d}_{2} = 20cm

\dfrac{10}{20} \times {( \dfrac{{20}^{2} - l^{2}}{{10}^{2} - l^{2}} )}^{2} = \dfrac{25}{2}

400 - l^{2} = 5( 100 - l^{2} )

4l^{2} = 100 ⇒ l^{2} = 25

l = 5cm

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