NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

Ans: Provided in the question,

Magnetic field strength \[B=0.25\,T\] 

Torque on the bar magnet, \[T=4.5\times {{10}^{-2}}J\] 

Angle between the given bar magnet and the external magnetic field, \[\theta =30{}^\circ \]

Torque is related to magnetic moment (M) as: 

\[T=MB\sin (\theta )\]

\[\Rightarrow M=\frac{4.5\times {{10}^{-2}}}{0.25\times \sin 30{}^\circ }=0.36J/T\]

Clearly, the magnetic moment of the magnet is \[0.36J/T\].

2. A short bar magnet of magnetic moment \[M=0.32\,J/T\] is placed in a uniform magnetic field of \[0.15\,T\]. If the bar is free to rotate in the plane of the field, which orientation and would correspond to its

a) Stable?

Ans: It is provided that moment of the bar magnet, \[M=0.32J/T\].

External magnetic field, \[B=0.15T\]

It is considered as being in stable equilibrium, when the bar magnet is aligned along the magnetic field. Therefore, the angle \[\theta \], between the bar magnet and the magnetic field is \[0{}^\circ \] .

Potential energy of the system \[=-MB\cos (\theta )\]

\[\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (0)=-4.8\times {{10}^{-2}}J\] 

Hence the potential energy is \[=-4.8\times {{10}^{-2}}J\]

b) Unstable equilibrium? What is the potential energy of the magnet in each case? 

Ans: It is provided that moment of the bar magnet, \[M=0.32J/T\]

External magnetic field, \[B=0.15T\]

When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium, \[\theta =180{}^\circ \]

Potential energy of the system is hence\[=-MB\cos (\theta )\]

\[\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (180{}^\circ )=4.8\times {{10}^{-2}}J\] 

Hence the potential energy is \[=4.8\times {{10}^{-2}}J\].

3. A closely wound solenoid of \[800\] turns and area of cross section \[2.5\times {{10}^{-4}}\,{{m}^{2}}\] carries a current of \[3.0A\]. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? 

Ans:  It is provided that number of turns in the solenoid, \[n=800\].

Area of cross-section, \[A=2.5\times {{10}^{-4}}{{m}^{2}}\]

Current in the solenoid, \[I=3.0A\]

A current-carrying solenoid is analogous to a bar magnet because a magnetic field develops along its axis, i.e., along its length joining the north and south poles.

The magnetic moment due to the given current-carrying solenoid is calculated as:

\[M=nIA=800\times 3\times 2.5\times {{10}^{-4}}=0.6J/T\]

Thus, the associated magnetic moment \[=0.6J/T\]

4. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of \[0.25\,T\] is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of \[30{}^\circ \] with the direction of applied field? 

Ans: Given is the magnetic field strength, \[B=0.25\,T\]

Magnetic moment, \[M=0.6\,/T\]

The angle, \[\theta \] between the axis of the turns of the solenoid and the direction of the external applied field is \[30{}^\circ \] .

Hence, the torque acting on the solenoid is given as: 

\[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =0.6\times 0.25\sin (30{}^\circ )\]

\[\Rightarrow \tau =7.5\times {{10}^{-2}}J\]

Hence the magnitude of torque is \[=7.5\times {{10}^{-2}}J\]

5. A bar magnet of magnetic moment \[1.5\,J/T\] lies aligned with the direction of a uniform magnetic field of \[0.22\,T\]. 

a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

Ans: Provided that,

Magnetic moment, \[M=1.5J/T\]

Magnetic field strength, \[B=0.22\,T\]

(i) Initial angle between the magnetic field and the axis is, \[{{\theta }_{1}}=0{}^\circ \]

Final angle between the magnetic field and the axis is, \[{{\theta }_{2}}=90{}^\circ \] 

The work that would be required to make the magnetic moment perpendicular to the direction of magnetic field would be:

\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]

\[\Rightarrow W=-1.5\times 0.22(\cos 90{}^\circ -\cos 0{}^\circ )\]

\[\Rightarrow W=-0.33(0-1)\]

\[\Rightarrow W=0.33\,J\]

(ii) Initial angle between the magnetic field and the axis, \[{{\theta }_{1}}=0{}^\circ \]

Final angle between the magnetic field and the axis, \[{{\theta }_{2}}=180{}^\circ \]

The work that would be required to make the magnetic moment opposite (180 degrees) to the direction of magnetic field is given as:

\[W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})\]

\[\Rightarrow W=-1.5\times 0.22(\cos 180{}^\circ -\cos 0{}^\circ )\]

\[\Rightarrow W=-0.33(-1-1)\]

\[\Rightarrow W=0.66J\]

b) What is the torque on the magnet in cases (i) and (ii)? 

Ans: For the first (i) case,

\[\theta ={{\theta }_{1}}=90{}^\circ \]

Hence the Torque, \[\vec{\tau }=\vec{M}\times \vec{B}\]

And its magnitude is:\[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =1.5\times 0.22\sin (90{}^\circ )\]

\[\Rightarrow \tau =0.33Nm\]

Hence the torque involved is \[=0.33Nm\]

For the second-(ii) case:

\[\theta ={{\theta }_{1}}=180{}^\circ \]

And its magnitude of the torque is:\[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =1.5\times 0.22\sin (180{}^\circ )\]

\[\Rightarrow \tau =0Nm\]

Hence the torque is zero.

6. A closely wound solenoid of \[2000\] turns and area of cross-section \[1.6\times {{10}^{-4}}{{m}^{2}}\], carrying a current of \[4.0\,A\], is suspended through its center allowing it to turn in a horizontal plane. 

a) What is the magnetic moment associated with the solenoid? 

Ans: Given is the number of turns on the solenoid, \[n=2000\]

Area of cross-section of the solenoid, \[A=1.6\times {{10}^{-4}}{{m}^{2}}\]

Current in the solenoid, \[I=4A\]

The magnetic moment inside the solenoid at the axis is calculated as:

\[M=nAI=2000\times 1.6\times {{10}^{-4}}\times 4=1.28A{{m}^{2}}\]

b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of \[7.5\times {{10}^{-2}}T\] is set up at an angle of \[30{}^\circ \] with the axis of the solenoid?

Ans: Provided that,

Magnetic field, \[B=7.5\times {{10}^{-2}}T\]

Angle between the axis and the magnetic field of the solenoid, \[\theta =30{}^\circ \]

Torque, \[\tau =MB\sin (\theta )\]

\[\Rightarrow \tau =1.28\times 7.5\times {{10}^{-2}}\sin (30{}^\circ )\]

\[\Rightarrow \tau =4.8\times {{10}^{-2}}Nm\]

Given the magnetic field is uniform, and the force on the solenoid is zero. The torque on the solenoid is \[4.8\times {{10}^{-2}}Nm\].

7. A short bar magnet has a magnetic moment of \[0.48\,J/T\]. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of \[10\,cm\] from the center of the magnet on

a) the axis, 

Ans: Provided that the magnetic moment of the given bar magnet, \[M\]is \[0.48J/T\]

Given distance, \[d=10cm=0.1m\]

The magnetic field at d-distance, from the centre of the magnet on the axis is given by the relation:

\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}\]

here,

\[{{\mu }_{0}}=\] Permeability of free space\[=4\pi \times {{10}^{-7}}Tm/A\]

Substituting these values, \[B\] becomes as follows:

\[\Rightarrow B=\frac{4\pi \times {{10}^{-7}}}{4\pi }\frac{2\times 0.48}{{{0.1}^{3}}}\]

\[\Rightarrow B=0.96\times {{10}^{-4}}T=0.96\,G\]

The magnetic field is \[0.96G\] along the South-North direction.

b) the equatorial lines (normal bisector) of the magnet.

Ans: The magnetic field at a point which is \[d=10cm=0.1m\] away on the equatorial of the magnet is given as:

\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{d}^{3}}}\]

\[\Rightarrow B=\frac{4\pi \times {{10}^{-7}}}{4\pi }\frac{0.48}{{{0.1}^{3}}}\]

\[\Rightarrow B=0.48\times {{10}^{-4}}T=0.48G\]

The magnetic field is \[0.48G\]along the North-South direction.

Overview of Deleted Syllabus for CBSE Class 12 Physics Magnetism and Matter

Conclusion

NCERT Class 12 Physics Chapter 5 Exercise Solutions on Magnetism and Matter provided by Vedantu explains an in-depth exploration of magnetic properties and their practical applications. The Chapter provides a comprehensive understanding of how different materials interact with magnetic fields and the fundamental principles governing these interactions. By exploring topics such as magnetization, magnetic intensity, and hysteresis, as well as Earth's magnetism, this chapter equips students with essential knowledge for both academic and practical applications in technology and scientific research. From previous year's question papers, typically around 3–4 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills. 

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