
Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed $20\,cm$ in $\tan \,A$ position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to
(A) $20\,cm$
(B) $20 \times {\left( 2 \right)^{1/3}}\,cm$
(C) $20 \times {\left( 2 \right)^{2/3}}\,cm$
(D) $20 \times {\left( 2 \right)^{3/3}}\,cm$
Answer
214.5k+ views
Hint: Use the condition for the no deflection between the magnets, frame the relation between the length and the distance of the magnet from the compass needle. Substitute the known values to find the value of the distance of the second magnet from the compass needle.
Useful formula:
The formula of the no deflection in the position of the $\tan \,A$ from the compass needle is
$\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}$
Where ${M_1}$ is the length of the first magnet, ${M_2}$ is the length of the first magnet, ${d_1}$ is the distance of the first magnet from the compass needle and ${d_2}$ is the distance of the second magnet from the compass needle.
Complete step by step solution:
It is given that the
The shorter magnet is placed at a distance from $\tan \,A$ , ${d_1} = 20\,cm$
The longer magnet is twice the length of the shorter magnet.
Using the formula of the no deflection,
$\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}$
By cancelling the similar terms on both sides,
$\dfrac{{{M_1}}}{{{d_1}^3}} = \dfrac{{{M_2}}}{{{d_2}^3}}$
By rearranging the terms in the above step, we get
$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{{d_1}^3}}{{{d_2}^3}}$
Substituting the known values in the above equation,
$\dfrac{1}{2} = \dfrac{{{{20}^3}}}{{{d_2}^3}}$
By simplifying the above values,
${d_{{2^{}}}}^3 = 4000$
Hence the value of the distance of the second magnet from the compass needle is obtained as follows.
${d_2}^3 = 20 \times {\left( 2 \right)^{1/3}}\,cm$
Thus the option (B) is correct.
Note: In the question, it is given that the Two short magnets have equal pole strengths but one is twice as long as the other. Hence in the above calculation, the value of the length of the first magnet is taken as $1$ and that of the second magnet is $2\left( 1 \right) = 2$.
Useful formula:
The formula of the no deflection in the position of the $\tan \,A$ from the compass needle is
$\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}$
Where ${M_1}$ is the length of the first magnet, ${M_2}$ is the length of the first magnet, ${d_1}$ is the distance of the first magnet from the compass needle and ${d_2}$ is the distance of the second magnet from the compass needle.
Complete step by step solution:
It is given that the
The shorter magnet is placed at a distance from $\tan \,A$ , ${d_1} = 20\,cm$
The longer magnet is twice the length of the shorter magnet.
Using the formula of the no deflection,
$\dfrac{{2{M_1}{\mu _0}}}{{4\pi {d_1}^3}} = \dfrac{{2{M_2}{\mu _0}}}{{4\pi {d_2}^3}}$
By cancelling the similar terms on both sides,
$\dfrac{{{M_1}}}{{{d_1}^3}} = \dfrac{{{M_2}}}{{{d_2}^3}}$
By rearranging the terms in the above step, we get
$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{{d_1}^3}}{{{d_2}^3}}$
Substituting the known values in the above equation,
$\dfrac{1}{2} = \dfrac{{{{20}^3}}}{{{d_2}^3}}$
By simplifying the above values,
${d_{{2^{}}}}^3 = 4000$
Hence the value of the distance of the second magnet from the compass needle is obtained as follows.
${d_2}^3 = 20 \times {\left( 2 \right)^{1/3}}\,cm$
Thus the option (B) is correct.
Note: In the question, it is given that the Two short magnets have equal pole strengths but one is twice as long as the other. Hence in the above calculation, the value of the length of the first magnet is taken as $1$ and that of the second magnet is $2\left( 1 \right) = 2$.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Collision: Meaning, Types & Examples in Physics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Atomic Structure: Definition, Models, and Examples

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Average and RMS Value in Physics: Formula, Comparison & Application

