Answer

Verified

420.3k+ views

**Hint :**When a bar magnet is placed in a magnetic field, the two poles of the bar magnet experience equal to the opposite forces. In other words, at one end nullifies the force at the other end. Hence, the bar magnet experiences only the torque due to the magnetic field and there is no force.

Torque is the measure of the force that can cause an object to rotate about an axis.

FORMULA USED:

$ \overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B} $

$ \overrightarrow{B}= $ Magnetic field

$ \overrightarrow{M}= $ Magnetic dipole moment

U = W

$ =MB\left( \cos {{\theta }_{2}}-\cos {{\theta }_{1}} \right) $

**Complete step by step answer**

When a magnetic dipole of moment M is held at an angle with the direction of a uniform magnetic field B, the magnitude of torque acting on the diploe is

$ \overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B} $

$ =MB\sin \theta $

This torque tends to align the dipole in the direction of the field. Work has to be done in rotating dipoles against the action of the torque. This work done is stored in the magnetic dipole as potential energy of the dipole.

Now, the small amount of work is done in rotating the dipole through a small angles $ d\theta $ is

$ dW=\tau d\theta $

$ =MB\sin \theta \cdot d\theta $

Total work done in rotating the dipole from $ \theta ={{\theta }_{1}}\text{ }to\text{ }\theta ={{\theta }_{2}}\text{ is:} $

$ W=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{MB\sin \theta \cdot d\theta } $

$ =MB\left| -\cos \theta \right|_{{{\theta }_{1}}}^{{{\theta }_{2}}} $

$ W=-MB\left[ \cos {{\theta }_{2}}-\cos {{\theta }_{1}} \right] $ ……... (1)

When

$ \text{ }{{\theta }_{1}}\text{ = 90}{}^\circ $ and $ {{\theta }_{2}}=\theta $

Then

U =W

$ =-MB(\cos \theta -\cos 90{}^\circ ) $

= $ U=-MB\cos \theta $

In vector form;

$ U=\overrightarrow{M}\centerdot \overrightarrow{B} $

When the magnet is in the stable equilibrium that is when the magnetic dipole is aligned along the magnetic field. The stable equilibrium has minimum potential energy.

When $ \theta =0{}^\circ $

$ U=-MB\cos 0{}^\circ $

$ U=-MB $ …..……... (2)

In our question the value of M and B are given as:

$ M=0.4J{{T}^{-1}} $

$ B=0.16T $

Put these values in equation (2)

$ U=-(0.4)\times (0.16) $

$ U=-0.064J $

**Therefore option (A) is correct.**

**Note**

Using the concept, we can also find out the value of U for unstable equilibrium, that is when $ \theta =180{}^\circ $ and expression becomes:

$ U=-MB\cos 180{}^\circ $

$ =-MB(-1) $

$ U=MB $

Having maximum potential energy.

Remember that when a bar magnet is placed in a uniform magnetic field it experiences only torque and no other force.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Choose the antonym of the word given below Furious class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE