
The magnetic induction at a point 1${A^o}$ away from a proton measured along its axis of spin is (magnetic moment of the proton is $1.4 \times {10^{ - 26}}A{m^2}$).
$
(a){\text{ 0}}{\text{.28mT}} \\
(b){\text{ 28mT}} \\
(c){\text{ 0}}{\text{.028mT}} \\
(d){\text{ 2}}{\text{.8mT}} \\
$
Answer
220.8k+ views
Hint: In this question use the direct formula that the magnetic induction at any point on the axis of magnetic dipoles is given as $B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2m}}{{{d^3}}}$. The basic conversion of $1{A^o} = 1 \times {10^{ - 10}}$ meter along with proper substitution of values in the direct formula will help approach the solution.
Complete step-by-step solution -
Given data:
Magnetic moment (m) of the proton = $1.4 \times {10^{ - 26}}A{m^2}$
Distance (d) of the point and the proton = 1${A^o}$
Now as we know that on the axis of magnetic dipoles, magnetic induction due to any point is given as
$B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2m}}{{{d^3}}}$......................... (1), often measured in Tesla (T).
Where, B = magnetic induction measured in TESLA.
${\mu _o}$ = permeability in the air.
m = magnetic moment of the proton.
d = distance of the point and the proton.
Now as we know that ${\mu _o}$ is the permeability in the air and whose value is $4\pi \times {10^{ - 7}}$(H/m).
And we all know that $1{A^o} = 1 \times {10^{ - 10}}$ meter.
Therefore, d = $1 \times {10^{ - 10}}$meter.
Now substitute the values in equation (1) we have,
$ \Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }} \times \dfrac{{2 \times \left( {1.4 \times {{10}^{ - 26}}} \right)}}{{{{\left( {1 \times {{10}^{ - 10}}} \right)}^3}}}$T
Now simplify this equation we have,
$ \Rightarrow B = {10^{ - 7}} \times \dfrac{{2.8 \times {{10}^{ - 26}}}}{{1 \times {{10}^{ - 30}}}} = 2.8 \times {10^{ - 7 - 26 + 30}}$
$ \Rightarrow B = 2.8 \times {10^{ - 3}}$T
Now as we know 1mT = 1$ \times {10^{ - 3}}$T.
$ \Rightarrow B = 2.8$mT.
So this is the required magnetic induction at a point 1${A^o}$ away from a proton measured along its axis of spin.
So this is the required answer.
Hence option (D) is the correct answer.
Note – Magnetic induction is general and has the definition that it refers to the process by which any object is actually magnetized whenever it is placed in an external magnetic field. This magnetizing property of any material depends upon the composition of the material. It is advised to remember the direct formula as it helps save a lot of time.
Complete step-by-step solution -
Given data:
Magnetic moment (m) of the proton = $1.4 \times {10^{ - 26}}A{m^2}$
Distance (d) of the point and the proton = 1${A^o}$
Now as we know that on the axis of magnetic dipoles, magnetic induction due to any point is given as
$B = \dfrac{{{\mu _o}}}{{4\pi }} \times \dfrac{{2m}}{{{d^3}}}$......................... (1), often measured in Tesla (T).
Where, B = magnetic induction measured in TESLA.
${\mu _o}$ = permeability in the air.
m = magnetic moment of the proton.
d = distance of the point and the proton.
Now as we know that ${\mu _o}$ is the permeability in the air and whose value is $4\pi \times {10^{ - 7}}$(H/m).
And we all know that $1{A^o} = 1 \times {10^{ - 10}}$ meter.
Therefore, d = $1 \times {10^{ - 10}}$meter.
Now substitute the values in equation (1) we have,
$ \Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }} \times \dfrac{{2 \times \left( {1.4 \times {{10}^{ - 26}}} \right)}}{{{{\left( {1 \times {{10}^{ - 10}}} \right)}^3}}}$T
Now simplify this equation we have,
$ \Rightarrow B = {10^{ - 7}} \times \dfrac{{2.8 \times {{10}^{ - 26}}}}{{1 \times {{10}^{ - 30}}}} = 2.8 \times {10^{ - 7 - 26 + 30}}$
$ \Rightarrow B = 2.8 \times {10^{ - 3}}$T
Now as we know 1mT = 1$ \times {10^{ - 3}}$T.
$ \Rightarrow B = 2.8$mT.
So this is the required magnetic induction at a point 1${A^o}$ away from a proton measured along its axis of spin.
So this is the required answer.
Hence option (D) is the correct answer.
Note – Magnetic induction is general and has the definition that it refers to the process by which any object is actually magnetized whenever it is placed in an external magnetic field. This magnetizing property of any material depends upon the composition of the material. It is advised to remember the direct formula as it helps save a lot of time.
Recently Updated Pages
[Awaiting input: Please provide the content from "Ask AI Response," "Competitor 1," and "Competitor 2," so I can perform the analysis and synthesize the requested metadata and headings.]

Young’s Double Slit Experiment Derivation Explained

A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

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

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Electromagnetic Waves and Their Importance

