
Two short magnets (having magnetic moments in the ratio $27:{\text{ }}8$. When placed on the opposite sides of a deflection magnetometer causes no deflection. If the distance of the weaker magnetic is $0.12\,m$ from the centre of deflection magnetometer, the distance of the stronger magnet from the centre will be:
A. $0.17$
B. $0.19$
C. $0.18$
D. $0.16$
Answer
164.1k+ views
Hint:The method of electrical measurements known as deflection uses the deflection of measuring the instrument's index to calculate the current or other element under consideration. It is distinct from and the inverse of the zero or null method. The null deflection method must be used to calculate the distance of the stronger magnet from the centre.
Formula used:
Relation between magnetic moment strength and distance is given by:
$\dfrac{M}{d^3}=\text{constant}$
Where, M is the magnetic moment of the magnet and d is the distance of a magnet from the magnetometer.
Complete step by step solution:
In the question, we have given the ratio of two short magnets is $27:{\text{ }}8$ and the distance of the weaker magnet from the centre of the deflection magnetometer is $0.12\,m$. As we know that $\tan \,A$ is the location of the deflection magnetometer.
A bar magnet with magnetic moment ${M_2} = 8$ and the distance of the weak magnet is ${d_2} = 0.12\,m$ from the magnetic needle's centre. The second bar magnet, with magnetic moment ${M_1} = 27$. The second magnet is adjusted to cancel out the deflection caused by the first magnet.
Let assume the distance of the strong magnet from the magnetic needle’s centre be ${d_2}$. Because the magnetic fields produced by the two bar magnets in the magnetic needle's centre are equal in magnitude but opposite in direction ${B_1} = {B_2}$, the null deflection is given by:
$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{d_1^3}}{{d_2^3}}$
Substitute the given information in the above formula, then we have:
$\dfrac{{27}}{8} = \dfrac{{d_1^3}}{{{{(0.12)}^3}}} \\$
$\Rightarrow {d_1} = \sqrt[3]{{\dfrac{{27 \times {{(0.12)}^3}}}{8}}} \\$
$\Rightarrow {d_1} = \dfrac{{3 \times 0.12}}{2} \\$
$\therefore {d_1} = 0.18\,m \\$
Therefore, the distance of the stronger magnet from the centre will be $0.18\,m$.
Thus, the correct option is C.
Note: It should be noted that we can be solved without even writing a single line of code. Simply remember that the magnetic moment is proportional to the distance cubed. As a result, the magnetic moment ratio will be the cube of the distance ratio given.
Formula used:
Relation between magnetic moment strength and distance is given by:
$\dfrac{M}{d^3}=\text{constant}$
Where, M is the magnetic moment of the magnet and d is the distance of a magnet from the magnetometer.
Complete step by step solution:
In the question, we have given the ratio of two short magnets is $27:{\text{ }}8$ and the distance of the weaker magnet from the centre of the deflection magnetometer is $0.12\,m$. As we know that $\tan \,A$ is the location of the deflection magnetometer.
A bar magnet with magnetic moment ${M_2} = 8$ and the distance of the weak magnet is ${d_2} = 0.12\,m$ from the magnetic needle's centre. The second bar magnet, with magnetic moment ${M_1} = 27$. The second magnet is adjusted to cancel out the deflection caused by the first magnet.
Let assume the distance of the strong magnet from the magnetic needle’s centre be ${d_2}$. Because the magnetic fields produced by the two bar magnets in the magnetic needle's centre are equal in magnitude but opposite in direction ${B_1} = {B_2}$, the null deflection is given by:
$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{d_1^3}}{{d_2^3}}$
Substitute the given information in the above formula, then we have:
$\dfrac{{27}}{8} = \dfrac{{d_1^3}}{{{{(0.12)}^3}}} \\$
$\Rightarrow {d_1} = \sqrt[3]{{\dfrac{{27 \times {{(0.12)}^3}}}{8}}} \\$
$\Rightarrow {d_1} = \dfrac{{3 \times 0.12}}{2} \\$
$\therefore {d_1} = 0.18\,m \\$
Therefore, the distance of the stronger magnet from the centre will be $0.18\,m$.
Thus, the correct option is C.
Note: It should be noted that we can be solved without even writing a single line of code. Simply remember that the magnetic moment is proportional to the distance cubed. As a result, the magnetic moment ratio will be the cube of the distance ratio given.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Charging and Discharging of Capacitor
