JEE Exam  >  JEE Notes  >  DPP: Daily Practice Problems for JEE Main & Advanced  >  DPP for JEE: Daily Practice Problems- Magnetism and Matter (Solutions)

Magnetism and Matter Practice Questions - DPP for JEE

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


1. (c) Initial magnetic moment of each magnet = m × l.
As is clear from Fig., S
1
 and N
2
 neutralize each other.
Effective distance between,
N
1
 and S
2
 = 
?  
2. (a) The time period of oscillation of a freely suspended magnet is given
by,
Thus, 
Given, T = 4 sec, ,
So, 
or,
or,
3. (b) t = MB sin ?
t = iAB sin 90º
? 
Also, A = 1/2 (BC) (AD)
Page 2


1. (c) Initial magnetic moment of each magnet = m × l.
As is clear from Fig., S
1
 and N
2
 neutralize each other.
Effective distance between,
N
1
 and S
2
 = 
?  
2. (a) The time period of oscillation of a freely suspended magnet is given
by,
Thus, 
Given, T = 4 sec, ,
So, 
or,
or,
3. (b) t = MB sin ?
t = iAB sin 90º
? 
Also, A = 1/2 (BC) (AD)
But  
?
4. (a) M = 60 Am
2
 = 1.2 × 10
–3
 Nm, B
H
 = 40 × 10
–6
 Wb/m
2
  
 1.2 × 10
–3
 = 60 × 40 × 10
–6
 sin ?
  = sin 30°
 ? = 30°
5. (b) Electro magnet should be amenable to magnetisation and 
demagnetization
? retentivity and coercivity should be low.
6. (b) where 
When the magnet is cut into three pieces the pole strength will remain
the same and
M.I. (I') = 
Page 3


1. (c) Initial magnetic moment of each magnet = m × l.
As is clear from Fig., S
1
 and N
2
 neutralize each other.
Effective distance between,
N
1
 and S
2
 = 
?  
2. (a) The time period of oscillation of a freely suspended magnet is given
by,
Thus, 
Given, T = 4 sec, ,
So, 
or,
or,
3. (b) t = MB sin ?
t = iAB sin 90º
? 
Also, A = 1/2 (BC) (AD)
But  
?
4. (a) M = 60 Am
2
 = 1.2 × 10
–3
 Nm, B
H
 = 40 × 10
–6
 Wb/m
2
  
 1.2 × 10
–3
 = 60 × 40 × 10
–6
 sin ?
  = sin 30°
 ? = 30°
5. (b) Electro magnet should be amenable to magnetisation and 
demagnetization
? retentivity and coercivity should be low.
6. (b) where 
When the magnet is cut into three pieces the pole strength will remain
the same and
M.I. (I') = 
We have, Magnetic moment (M)
= Pole strength (m) × l
?  New magnetic moment,
? .
7. (a) Given that : B
1
 = 1.2 × 10
–2
 T, orientation of dipole with the field B
1
,
?
1
 = 15°
Hence, orientation of dipole with B
2
,
?
2
 = 60° – 15° = 45° (figure)
As the dipole is in equilibrium, therefore, the torque on the dipole due to
the two fields must be equal and opposite. If M be the magnetic
dipole moment of the dipole, then
t
1
 = t
2
or MB
1
 sin ?
1
 = MB
2
 sin ?
2
or,     B
2
 = 
=  = 4.4 × 10
–3
 Tesla
8. (a) Here, 
At neutral point,
Page 4


1. (c) Initial magnetic moment of each magnet = m × l.
As is clear from Fig., S
1
 and N
2
 neutralize each other.
Effective distance between,
N
1
 and S
2
 = 
?  
2. (a) The time period of oscillation of a freely suspended magnet is given
by,
Thus, 
Given, T = 4 sec, ,
So, 
or,
or,
3. (b) t = MB sin ?
t = iAB sin 90º
? 
Also, A = 1/2 (BC) (AD)
But  
?
4. (a) M = 60 Am
2
 = 1.2 × 10
–3
 Nm, B
H
 = 40 × 10
–6
 Wb/m
2
  
 1.2 × 10
–3
 = 60 × 40 × 10
–6
 sin ?
  = sin 30°
 ? = 30°
5. (b) Electro magnet should be amenable to magnetisation and 
demagnetization
? retentivity and coercivity should be low.
6. (b) where 
When the magnet is cut into three pieces the pole strength will remain
the same and
M.I. (I') = 
We have, Magnetic moment (M)
= Pole strength (m) × l
?  New magnetic moment,
? .
7. (a) Given that : B
1
 = 1.2 × 10
–2
 T, orientation of dipole with the field B
1
,
?
1
 = 15°
Hence, orientation of dipole with B
2
,
?
2
 = 60° – 15° = 45° (figure)
As the dipole is in equilibrium, therefore, the torque on the dipole due to
the two fields must be equal and opposite. If M be the magnetic
dipole moment of the dipole, then
t
1
 = t
2
or MB
1
 sin ?
1
 = MB
2
 sin ?
2
or,     B
2
 = 
=  = 4.4 × 10
–3
 Tesla
8. (a) Here, 
At neutral point,
= 5 ab-amp cm.
9. (a) ?
d
 < ?
p
 < ?
f
For diamagnetic substance ?
d
 is small and negative (10
–5
)
For paramagnetic substances ?
p
 is small and positive (10
–3
 to 10
–5
)
For ferromagnetic substanes ?
f
 is very large (10
3
 to 10
5
)
10. (d) PQ
6
 corresponds to the lowest potential energy among all the
configurations shown.
11. (d) tan   
?  
V = 6 × 10
–5
 T
H = 
?  B
total 
=
= 10 × 10
–5
 = 10
–4
T.
12. (b) Ferromagnetic substance has magnetic domains whereas
paramagnetic substances have magnetic dipoles which get attracted
to a magnetic field.
Diamagnetic substances do not have magnetic dipole but in the presence
of external magnetic field due to their orbital motion of electrons
these substances are repelled.
13. (c) Initially magnetic moment of system
 and moment of inertia
Page 5


1. (c) Initial magnetic moment of each magnet = m × l.
As is clear from Fig., S
1
 and N
2
 neutralize each other.
Effective distance between,
N
1
 and S
2
 = 
?  
2. (a) The time period of oscillation of a freely suspended magnet is given
by,
Thus, 
Given, T = 4 sec, ,
So, 
or,
or,
3. (b) t = MB sin ?
t = iAB sin 90º
? 
Also, A = 1/2 (BC) (AD)
But  
?
4. (a) M = 60 Am
2
 = 1.2 × 10
–3
 Nm, B
H
 = 40 × 10
–6
 Wb/m
2
  
 1.2 × 10
–3
 = 60 × 40 × 10
–6
 sin ?
  = sin 30°
 ? = 30°
5. (b) Electro magnet should be amenable to magnetisation and 
demagnetization
? retentivity and coercivity should be low.
6. (b) where 
When the magnet is cut into three pieces the pole strength will remain
the same and
M.I. (I') = 
We have, Magnetic moment (M)
= Pole strength (m) × l
?  New magnetic moment,
? .
7. (a) Given that : B
1
 = 1.2 × 10
–2
 T, orientation of dipole with the field B
1
,
?
1
 = 15°
Hence, orientation of dipole with B
2
,
?
2
 = 60° – 15° = 45° (figure)
As the dipole is in equilibrium, therefore, the torque on the dipole due to
the two fields must be equal and opposite. If M be the magnetic
dipole moment of the dipole, then
t
1
 = t
2
or MB
1
 sin ?
1
 = MB
2
 sin ?
2
or,     B
2
 = 
=  = 4.4 × 10
–3
 Tesla
8. (a) Here, 
At neutral point,
= 5 ab-amp cm.
9. (a) ?
d
 < ?
p
 < ?
f
For diamagnetic substance ?
d
 is small and negative (10
–5
)
For paramagnetic substances ?
p
 is small and positive (10
–3
 to 10
–5
)
For ferromagnetic substanes ?
f
 is very large (10
3
 to 10
5
)
10. (d) PQ
6
 corresponds to the lowest potential energy among all the
configurations shown.
11. (d) tan   
?  
V = 6 × 10
–5
 T
H = 
?  B
total 
=
= 10 × 10
–5
 = 10
–4
T.
12. (b) Ferromagnetic substance has magnetic domains whereas
paramagnetic substances have magnetic dipoles which get attracted
to a magnetic field.
Diamagnetic substances do not have magnetic dipole but in the presence
of external magnetic field due to their orbital motion of electrons
these substances are repelled.
13. (c) Initially magnetic moment of system
 and moment of inertia
I
1
 = I + I = 2I.
Finally when one of the magnet is removed then
M
2
 = M and I
2
 = I
So, 
14. (b) Force on a charged particle is given by F = qvB. Here v = 0 and
also resultant B is zero.
? Force = 0
15. (b)
 = 
= ,  
16. (d) d
1
 = 40°, d
2
 = 30°, d = ?
 
Read More
174 docs

Top Courses for JEE

174 docs
Download as PDF
Explore Courses for JEE exam

Top Courses for JEE

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Summary

,

past year papers

,

ppt

,

study material

,

Previous Year Questions with Solutions

,

Sample Paper

,

pdf

,

video lectures

,

Important questions

,

Semester Notes

,

Magnetism and Matter Practice Questions - DPP for JEE

,

Objective type Questions

,

Viva Questions

,

Magnetism and Matter Practice Questions - DPP for JEE

,

MCQs

,

mock tests for examination

,

practice quizzes

,

Extra Questions

,

Free

,

Magnetism and Matter Practice Questions - DPP for JEE

,

shortcuts and tricks

,

Exam

;