Vidaara.orgClass 12 · Physics
CodeVID-P12-02-MAG-01
Magnetism & Matter — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Take $\frac{\mu_0}{4\pi}=10^{-7}$ T m/A. Draw neat labelled diagrams wherever asked. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The magnetic moment of a dipole points from:
- A.N to S
- B.S to N
- C.east to west
- D.up to down
2.
At the magnetic poles the angle of dip is:
- A.0 degrees
- B.45 degrees
- C.90 degrees
- D.30 degrees
3.
Susceptibility of a diamagnetic material is:
- A.large positive
- B.small positive
- C.small negative
- D.zero
4.
Paramagnetic susceptibility varies with temperature as:
- A.$\chi\propto T$
- B.$\chi\propto 1/T$
- C.$\chi\propto T^2$
- D.independent of T
5.
The area of a hysteresis loop represents:
- A.retentivity
- B.energy lost per cycle
- C.coercivity
- D.magnetic moment
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define declination and dip.
7.
Write the torque and potential energy of a dipole in a field.
8.
Define retentivity and coercivity.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Compare diamagnetic, paramagnetic and ferromagnetic materials.
10.
At a place $B_H=0.26$ G and dip is $30^\circ$. Find the total field.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Draw and explain the hysteresis loop of a ferromagnetic material. Define retentivity and coercivity, and explain why soft iron is used for transformer cores and steel for permanent magnets.
Answer Key
Section A — Multiple Choice Questions
- (B) S to N
- (C) 90 degrees
- (C) small negative
- (B) $\chi\propto 1/T$
- (B) energy lost per cycle
Section B — Short Answer (2 marks)
- Declination: angle between geographic and magnetic north; dip: angle of total field with the horizontal.
- $\tau=mB\sin\theta$; $U=-mB\cos\theta$.
- Retentivity: $B$ left when $H=0$; coercivity: reverse $H$ needed to make $B=0$.
Section C — Short Answer (3 marks)
- Dia: $\chi<0$; para: small $\chi>0$ ($\chi\propto1/T$); ferro: large $\chi>0$.
- $B=\frac{B_H}{\cos I}=\frac{0.26}{\cos30^\circ}=0.30$ G.
Section D — Long Answer (5 marks)
- Loop shows lag of B behind H; soft iron has a thin loop (low loss) for cores; steel has a wide loop (high retentivity) for permanent magnets.
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