
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
205.2k+ 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
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

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
Atomic Structure: Definition, Models, and Examples

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Electric Field Due to a Uniformly Charged Ring Explained

Centre of Mass of Hollow and Solid Hemisphere Explained

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Other Pages
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Diffraction of Light - Young’s Single Slit Experiment

Wheatstone Bridge Explained: Principle, Working, and Uses

Charging and Discharging of Capacitor Explained

Electric field due to uniformly charged sphere class 12 physics JEE_Main

