Answer
64.8k+ views
Hint: To solve the asked question you need to know the formula of force between two poles separated by a distance $r$ , this formula is very frequently used and includes the use of constant ${\mu _o}$ which is numerically equal to $4\pi \times {10^{ - 7}}$ . Remember that the force between two given poles decreases as the distance between them increases, hence, the term $r$ always comes in denominator in the relation.
Complete step by step answer:
As explained in the hint section of the solution to the asked question, the best approach is to simply use the formula that gives us the value of force between two poles separated by a distance $r$ , and gives the relation between force, pole strengths of the two poles and the distance between the given two poles. This formula is given as:
$F = \dfrac{{{\mu _o}}}{{4\pi }}\left( {\dfrac{{{m_1}{m_2}}}{{{r^2}}}} \right)$
Where, ${\mu _o}$ is the permeability constant whose value is given as $4\pi \times {10^{ - 7}}$
${m_1}$ is the poles strength of first pole,
${m_2}$ is the pole strength of the other pole and,
$r$ is the distance by which the two given poles are separated.
The question has already given us the pole strengths of north pole and south pole as:
$
\Rightarrow {m_N} = 6 \times {10^{ - 3}}\,Am \\
\Rightarrow {m_S} = 8 \times {10^{ - 3}}\,Am \\
$
The distance between the two given poles is given to us to be:
$r = 10\,cm$
Or, $r = 0.1\,m$
Now, all we need to do is to simply substitute the values of pole strength and distance between the poles to find out the answer using the above-mentioned formula.
Hence, for our case, we can write:
$F = \dfrac{{{\mu _o}}}{{4\pi }}\left( {\dfrac{{{m_N}{m_S}}}{{{r^2}}}} \right)$
Now, let’s substitute the values to find the answer as:
\[\Rightarrow F = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }}\left( {\dfrac{{6 \times {{10}^{ - 3}} \times \,8 \times {{10}^{ - 3}}}}{{{{\left( {0.1} \right)}^2}}}} \right)\]
Upon further solving and cancelling out terms, we get the final value of force between the two poles separated by the given distance as:
$\Rightarrow F = 48 \times {10^{ - 11}}{\kern 1pt} N$
We can see that our answer matches the value with the given value in option (C). Hence, option (C) is the correct answer to the question.
Note: Many students make an error by trying to find the force between them using the formula or relation of dipole moment, which is completely wrong as although that formula is very commonly used, it is not what is being asked in the question.
Complete step by step answer:
As explained in the hint section of the solution to the asked question, the best approach is to simply use the formula that gives us the value of force between two poles separated by a distance $r$ , and gives the relation between force, pole strengths of the two poles and the distance between the given two poles. This formula is given as:
$F = \dfrac{{{\mu _o}}}{{4\pi }}\left( {\dfrac{{{m_1}{m_2}}}{{{r^2}}}} \right)$
Where, ${\mu _o}$ is the permeability constant whose value is given as $4\pi \times {10^{ - 7}}$
${m_1}$ is the poles strength of first pole,
${m_2}$ is the pole strength of the other pole and,
$r$ is the distance by which the two given poles are separated.
The question has already given us the pole strengths of north pole and south pole as:
$
\Rightarrow {m_N} = 6 \times {10^{ - 3}}\,Am \\
\Rightarrow {m_S} = 8 \times {10^{ - 3}}\,Am \\
$
The distance between the two given poles is given to us to be:
$r = 10\,cm$
Or, $r = 0.1\,m$
Now, all we need to do is to simply substitute the values of pole strength and distance between the poles to find out the answer using the above-mentioned formula.
Hence, for our case, we can write:
$F = \dfrac{{{\mu _o}}}{{4\pi }}\left( {\dfrac{{{m_N}{m_S}}}{{{r^2}}}} \right)$
Now, let’s substitute the values to find the answer as:
\[\Rightarrow F = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }}\left( {\dfrac{{6 \times {{10}^{ - 3}} \times \,8 \times {{10}^{ - 3}}}}{{{{\left( {0.1} \right)}^2}}}} \right)\]
Upon further solving and cancelling out terms, we get the final value of force between the two poles separated by the given distance as:
$\Rightarrow F = 48 \times {10^{ - 11}}{\kern 1pt} N$
We can see that our answer matches the value with the given value in option (C). Hence, option (C) is the correct answer to the question.
Note: Many students make an error by trying to find the force between them using the formula or relation of dipole moment, which is completely wrong as although that formula is very commonly used, it is not what is being asked in the question.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)