
A magnet of magnetic moment \[M\] is situated with its axis along the direction of a magnetic field of strength\[B\]. The work done in rotating it by an angle of \[{180^0}\] will be:
A. \[ - MB\]
B. \[ + MB\]
C. \[ + 2MB\]
D. \[Zero\]
Answer
164.1k+ views
Hint:
To solve this question we have to use the basic formula of work done in moving a dipole in an external magnetic field. Use the given data and by putting it into the equation we can directly solve the question.
Formula used:
\[W = MB(1 - \cos \theta )\]
\[M\]- magnetic moment of the dipole
\[B\]- magnetic field strength
Complete step by step solution:
let us solve the given question by using the given data.
Given data: \[M\]- magnetic moment of the dipole.
\[B\]- magnetic field strength
\[\theta = {180^0}\]
Now, by using formula for work done in moving the dipole with magnetic moment in given magnetic field by the angle of \[{180^0}\] we have:
\[W = MB(1 - \cos \theta )\]
By using \[\theta = {180^0}\]in above equation we get
\[ \Rightarrow W = MB(1 - \cos {180^0})\]
\[ \Rightarrow W = MB(1 - ( - 1))\]
\[ \Rightarrow W = 2MB\]
Hence, the work done on the magnet to rotate it with the angle of \[{180^0}\]is \[2MB\].
Correct answer is option c.
Therefore, the correct option is C.
Note:
In this question, the dipole is along the magnetic field also called stable equilibrium position and rotating the magnet by \[{180^0}\] then it will be at unstable equilibrium position. A particle always tries to remain at a stable equilibrium position so whenever we move the dipole from its stable equilibrium position we have to do some extra work.
To solve this question we have to use the basic formula of work done in moving a dipole in an external magnetic field. Use the given data and by putting it into the equation we can directly solve the question.
Formula used:
\[W = MB(1 - \cos \theta )\]
\[M\]- magnetic moment of the dipole
\[B\]- magnetic field strength
Complete step by step solution:
let us solve the given question by using the given data.
Given data: \[M\]- magnetic moment of the dipole.
\[B\]- magnetic field strength
\[\theta = {180^0}\]
Now, by using formula for work done in moving the dipole with magnetic moment in given magnetic field by the angle of \[{180^0}\] we have:
\[W = MB(1 - \cos \theta )\]
By using \[\theta = {180^0}\]in above equation we get
\[ \Rightarrow W = MB(1 - \cos {180^0})\]
\[ \Rightarrow W = MB(1 - ( - 1))\]
\[ \Rightarrow W = 2MB\]
Hence, the work done on the magnet to rotate it with the angle of \[{180^0}\]is \[2MB\].
Correct answer is option c.
Therefore, the correct option is C.
Note:
In this question, the dipole is along the magnetic field also called stable equilibrium position and rotating the magnet by \[{180^0}\] then it will be at unstable equilibrium position. A particle always tries to remain at a stable equilibrium position so whenever we move the dipole from its stable equilibrium position we have to do some extra work.
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