Answer
416.7k+ views
Hint:Use the below formula of the magnetic moment and substitute the values known to find the value of the moment. Then substitute this and also the other required values in the work done formula, to find the work done to move the magnet from the stable to unstable position.
Formula used:
(1) The formula of the magnetic moment is given by
$M = \dfrac{T}{{B\sin \theta }}$
Where $M$ is the magnetic moment, $T$ is the torque experienced by the magnet, $B$ is the magnetic field and $\theta $ is the angle between the axis of the magnet with the external magnetic field.
(2) The formula of the work done to move the magnet from a stable to the unstable position is given by
$W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)$
Where $W$ is the work done to move the magnet.
Complete step by step solution:
It is given that the
The magnetic field, $B = 600\,G$
The angle that the axis makes with the magnetic field, $\theta = {30^ \circ }$
The torque experienced by the magnet, $T = 0.012\,Nm$
Using the formula of the magnetic moment,
$M = \dfrac{T}{{B\sin \theta }}$
Substituting the known values in the above formula,
$M = \dfrac{{0.012}}{{600 \times {{10}^{ - 4}} \times \sin {{30}^ \circ }}}$
By performing the various arithmetic an the trigonometric operation, we get
$M = 0.4\,A{m^2}$
Hence the magnetic moment is obtained as $0.4\,A{m^2}$ .
Using the formula (2) to calculate the work done,
$W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)$
Substituting the known parameters in it, we get
$W = - 0.4 \times 600 \times {10^{ - 4}}\left( {\cos {{180}^ \circ } - \cos {0^ \circ }} \right)$
By simplifying the above equation, we get
$W = 0.048\,J$
Hence the work done to move the magnet from the stable to the unstable position is $0.048\,J$ .
Note:In the above solution, the value of the one Gauss is substituted as $1 \times {10^{ - 4}}$ Tesla. This is because the derived $SI$ unit of the magnetic moment and it is derived and formed as the unit of $Wb{m^{ - 2}}$ . The most stable position is taken at ${180^0}$ , this is because the magnet has maximum potential energy at this angle and minimum at ${0^ \circ }$ .
Formula used:
(1) The formula of the magnetic moment is given by
$M = \dfrac{T}{{B\sin \theta }}$
Where $M$ is the magnetic moment, $T$ is the torque experienced by the magnet, $B$ is the magnetic field and $\theta $ is the angle between the axis of the magnet with the external magnetic field.
(2) The formula of the work done to move the magnet from a stable to the unstable position is given by
$W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)$
Where $W$ is the work done to move the magnet.
Complete step by step solution:
It is given that the
The magnetic field, $B = 600\,G$
The angle that the axis makes with the magnetic field, $\theta = {30^ \circ }$
The torque experienced by the magnet, $T = 0.012\,Nm$
Using the formula of the magnetic moment,
$M = \dfrac{T}{{B\sin \theta }}$
Substituting the known values in the above formula,
$M = \dfrac{{0.012}}{{600 \times {{10}^{ - 4}} \times \sin {{30}^ \circ }}}$
By performing the various arithmetic an the trigonometric operation, we get
$M = 0.4\,A{m^2}$
Hence the magnetic moment is obtained as $0.4\,A{m^2}$ .
Using the formula (2) to calculate the work done,
$W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)$
Substituting the known parameters in it, we get
$W = - 0.4 \times 600 \times {10^{ - 4}}\left( {\cos {{180}^ \circ } - \cos {0^ \circ }} \right)$
By simplifying the above equation, we get
$W = 0.048\,J$
Hence the work done to move the magnet from the stable to the unstable position is $0.048\,J$ .
Note:In the above solution, the value of the one Gauss is substituted as $1 \times {10^{ - 4}}$ Tesla. This is because the derived $SI$ unit of the magnetic moment and it is derived and formed as the unit of $Wb{m^{ - 2}}$ . The most stable position is taken at ${180^0}$ , this is because the magnet has maximum potential energy at this angle and minimum at ${0^ \circ }$ .
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)