
The distance of two points on the axis of a magnet from its centre are 10 cm and 20 cm respectively. The ratio of magnetic intensity at these points is 12.5 : 1. The length of the magnet will be
( a) 5 cm
(b) 25 cm
(c) 10 cm
( d) 20 cm
Answer
233.1k+ views
Hint:
In this question, we have given the distance of two points on the axis of magnet from its centre and ratio of magnetic intensity is also given. We have to find out the length of the magnet. First we use the formula of magnetic intensity at the axial point to solve this question . Now we put the values of r and B and by solving the equation, we get the length of the magnet.
Formula used:
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
Complete step by step solution:
We know the magnetic intensity on the axis of the magnet from its centre at a distance r is
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
The ratio of magnetic intensity on the axis at a distance ${{r}_{1}}$and ${{r}_{2}}$is ${{B}_{1}}$ and ${{B}_{2}}$
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{12.5}{1}$
Given ${{r}_{1}}$= 10 cm and ${{r}_{2}}$= 20 cm
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{\dfrac{{{u}_{0}}2{{r}_{1}}M}{4\pi {{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}}{\dfrac{{{u}_{0}}2{{r}_{2}}M}{4\pi {{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}}$
Which is equal to $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{{{r}_{1}}{{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}{{{r}_{2}}{{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}$
now we put the values in the above equation and we get
$\dfrac{12.5}{1}=\dfrac{10{{({{20}^{2}}-{{L}^{2}})}^{2}}}{20{{({{10}^{2}}-{{L}^{2}})}^{2}}}$
Hence by solving it, we get $\dfrac{5}{1}=\dfrac{{{20}^{2}}-{{L}^{2}}}{{{10}^{2}}-{{L}^{2}}}$
By cross multiplying, we get
$500-5{{L}^{2}}=400-{{L}^{2}}$
That is $4{{L}^{2}}=100$
Hence, we get L = 5 cm
We know length of the magnet is 2L
Then 2L = $2\times 5$ cm
L = 10 cm
Thus, Option (C ) is correct.
Note:
Students make mistakes while comparing the equations. We should take care while comparing the equations to get the correct answer. Also recall the complete formula of axial and equatorial magnetic field as they can be used in such type of questions.
In this question, we have given the distance of two points on the axis of magnet from its centre and ratio of magnetic intensity is also given. We have to find out the length of the magnet. First we use the formula of magnetic intensity at the axial point to solve this question . Now we put the values of r and B and by solving the equation, we get the length of the magnet.
Formula used:
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
Complete step by step solution:
We know the magnetic intensity on the axis of the magnet from its centre at a distance r is
$B=\dfrac{{{u}_{0}}2rM}{4\pi {{({{r}^{2}}-{{L}^{2}})}^{2}}}$
The ratio of magnetic intensity on the axis at a distance ${{r}_{1}}$and ${{r}_{2}}$is ${{B}_{1}}$ and ${{B}_{2}}$
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{12.5}{1}$
Given ${{r}_{1}}$= 10 cm and ${{r}_{2}}$= 20 cm
Thus $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{\dfrac{{{u}_{0}}2{{r}_{1}}M}{4\pi {{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}}{\dfrac{{{u}_{0}}2{{r}_{2}}M}{4\pi {{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}}$
Which is equal to $\dfrac{{{B}_{1}}}{{{B}_{2}}}=\dfrac{{{r}_{1}}{{({{r}_{2}}^{2}-{{L}^{2}})}^{2}}}{{{r}_{2}}{{({{r}_{1}}^{2}-{{L}^{2}})}^{2}}}$
now we put the values in the above equation and we get
$\dfrac{12.5}{1}=\dfrac{10{{({{20}^{2}}-{{L}^{2}})}^{2}}}{20{{({{10}^{2}}-{{L}^{2}})}^{2}}}$
Hence by solving it, we get $\dfrac{5}{1}=\dfrac{{{20}^{2}}-{{L}^{2}}}{{{10}^{2}}-{{L}^{2}}}$
By cross multiplying, we get
$500-5{{L}^{2}}=400-{{L}^{2}}$
That is $4{{L}^{2}}=100$
Hence, we get L = 5 cm
We know length of the magnet is 2L
Then 2L = $2\times 5$ cm
L = 10 cm
Thus, Option (C ) is correct.
Note:
Students make mistakes while comparing the equations. We should take care while comparing the equations to get the correct answer. Also recall the complete formula of axial and equatorial magnetic field as they can be used in such type of questions.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

