Solution of Magnetic Effects of Current Test 03 Class 12 Notes | EduRev

Class 12 : Solution of Magnetic Effects of Current Test 03 Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-03 
CLASS - XII PHYSICS (Magnetic effects of current and magnetism) 
[ANSWERS] 
Ans 1. Intensity of magnetization increases with the increase in applied magnetic field. 
Ans 2. Curie temperature of iron is about 770
0
C but when it is heated to a very higher 
temperature magnetism of iron further gets lost and it will not retain magnetism. 
Ans 3. Magnetic field due to semicircle QR at C. is 
1
1
1 0 2
 =  
2
I
B
R
µ p
µp
Magnetic field due to semicircle is at C is 
2
2
1 0 2
 =  
2
I
B
R
µ p
µp
 Net field     B  =  B 1 –  B 2 
1 2
1 2
1 0 1 1
 =  2 I  -
2 4
0 1 1
 =    -
4
B
R R
I
B
R R
µ
p
p
µ
? ?
? ?
? ?
? ?
? ?
? ?
Ans 4. m = NIA 
m = 1× 10
-4
× 2
m = 2 × 10
-4
 Am
2
.
Ans 5.   (a)  B = 0.10T 
? = 0
0
     (Normal to plane of the coil)
 I = 5.0 A , Area = 10
-5
 m2      n = 10
29
 m
-3
 
0
sin
sin 0 0
MB
MB
t ?
t
=
= =
 (b) Total force on the coil  =  0 Newton 
 (c) Fav  =  q (vd
 
B ×
 
) 
( I = neAVd) 
Fav  = 
qI
B
neA
× 
Fav = 
29 5
5 0.10
 
10 10
IB
nA
- ×
=
×
25
5 10 Fav N
- = ×
Page 2


CBSE TEST PAPER-03 
CLASS - XII PHYSICS (Magnetic effects of current and magnetism) 
[ANSWERS] 
Ans 1. Intensity of magnetization increases with the increase in applied magnetic field. 
Ans 2. Curie temperature of iron is about 770
0
C but when it is heated to a very higher 
temperature magnetism of iron further gets lost and it will not retain magnetism. 
Ans 3. Magnetic field due to semicircle QR at C. is 
1
1
1 0 2
 =  
2
I
B
R
µ p
µp
Magnetic field due to semicircle is at C is 
2
2
1 0 2
 =  
2
I
B
R
µ p
µp
 Net field     B  =  B 1 –  B 2 
1 2
1 2
1 0 1 1
 =  2 I  -
2 4
0 1 1
 =    -
4
B
R R
I
B
R R
µ
p
p
µ
? ?
? ?
? ?
? ?
? ?
? ?
Ans 4. m = NIA 
m = 1× 10
-4
× 2
m = 2 × 10
-4
 Am
2
.
Ans 5.   (a)  B = 0.10T 
? = 0
0
     (Normal to plane of the coil)
 I = 5.0 A , Area = 10
-5
 m2      n = 10
29
 m
-3
 
0
sin
sin 0 0
MB
MB
t ?
t
=
= =
 (b) Total force on the coil  =  0 Newton 
 (c) Fav  =  q (vd
 
B ×
 
) 
( I = neAVd) 
Fav  = 
qI
B
neA
× 
Fav = 
29 5
5 0.10
 
10 10
IB
nA
- ×
=
×
25
5 10 Fav N
- = ×
Ans 6. A particle of mass (m) and change (q) moving with velocity? normal to B
 
 describes 
a circular path if      
 
2
0
2
 qBvsin   ( =90 )
=>  qvB
 => (1)
mv
r
mv
r
mv
r
Bq
? ? = ?
=
= - - - - Since Time period of Revolution 
During circular path = 
Circumference of circle
velocity
=> 
2 r
T
v
p
= ( .(1)
Bqr
v from eg
m
? = ) 
=>    T = 
2 r m
Bqr
p ×
2
 -----(2)
m
T
Bq
p
=
Kinetic energy K.E = 
2
1
 mv
2
=> KE = 
2
1
 
2
Bqr
m
m
? ?
? ?
? ?
2 2 2
(3)
2
B q r
KE
m
= - - - 
Ans 7. If n be the no, of turns per unit length I be the current flowing through the Toroid 
 Then from Ampere’s circuital law 
Page 3


CBSE TEST PAPER-03 
CLASS - XII PHYSICS (Magnetic effects of current and magnetism) 
[ANSWERS] 
Ans 1. Intensity of magnetization increases with the increase in applied magnetic field. 
Ans 2. Curie temperature of iron is about 770
0
C but when it is heated to a very higher 
temperature magnetism of iron further gets lost and it will not retain magnetism. 
Ans 3. Magnetic field due to semicircle QR at C. is 
1
1
1 0 2
 =  
2
I
B
R
µ p
µp
Magnetic field due to semicircle is at C is 
2
2
1 0 2
 =  
2
I
B
R
µ p
µp
 Net field     B  =  B 1 –  B 2 
1 2
1 2
1 0 1 1
 =  2 I  -
2 4
0 1 1
 =    -
4
B
R R
I
B
R R
µ
p
p
µ
? ?
? ?
? ?
? ?
? ?
? ?
Ans 4. m = NIA 
m = 1× 10
-4
× 2
m = 2 × 10
-4
 Am
2
.
Ans 5.   (a)  B = 0.10T 
? = 0
0
     (Normal to plane of the coil)
 I = 5.0 A , Area = 10
-5
 m2      n = 10
29
 m
-3
 
0
sin
sin 0 0
MB
MB
t ?
t
=
= =
 (b) Total force on the coil  =  0 Newton 
 (c) Fav  =  q (vd
 
B ×
 
) 
( I = neAVd) 
Fav  = 
qI
B
neA
× 
Fav = 
29 5
5 0.10
 
10 10
IB
nA
- ×
=
×
25
5 10 Fav N
- = ×
Ans 6. A particle of mass (m) and change (q) moving with velocity? normal to B
 
 describes 
a circular path if      
 
2
0
2
 qBvsin   ( =90 )
=>  qvB
 => (1)
mv
r
mv
r
mv
r
Bq
? ? = ?
=
= - - - - Since Time period of Revolution 
During circular path = 
Circumference of circle
velocity
=> 
2 r
T
v
p
= ( .(1)
Bqr
v from eg
m
? = ) 
=>    T = 
2 r m
Bqr
p ×
2
 -----(2)
m
T
Bq
p
=
Kinetic energy K.E = 
2
1
 mv
2
=> KE = 
2
1
 
2
Bqr
m
m
? ?
? ?
? ?
2 2 2
(3)
2
B q r
KE
m
= - - - 
Ans 7. If n be the no, of turns per unit length I be the current flowing through the Toroid 
 Then from Ampere’s circuital law 
( )
( )
( )
( )
0
0
2
0
0
0
2
0
0
0
. d  =     total current flowing in the toroid
. d  =   2
cos 0  =   2
  =   2
. 2 r  =   2
 B  =  
r
r
B
B rnI
Bd rnI
B d rnI
B rnI
onI
p
p
µ
µ p
µ p
µ p
p µ p
µ
×
?
?
?
?
  
l
  
l
l
l


Ans 8. Force experienced by the charged particle moving at right angles to uniform magnetic 
field B
 
 with velocity v

 is given by   F

 = q ( v

 B ×
 
)  
 Initially Dee D 1 is negatively charged and 
Dee D 2 is positively charged so, the positive 
ion will get accelerated towards Dee D 1 
since the magnetic field is uniform and 
acting at right angles to the plane of the 
Dees so the ion completes a circular path in 
D1 when ions comes out into the gap, 
polarity of the Dee’s gets reversed used the 
ion is further accelerated towards Dee D 2 
with greater speed and cover a bigger 
semicircular path. This process is separated 
time and again and the speed of the ion 
becomes faster till it reaches the periphery 
of the dees where it is brought out by means of a deflecting plate and is made to 
bombard the target. 
 Since F = qVBsin90
0
 provides the necessary centripetal force to the ion to cover a 
circular path so we can say 
2
mv
qvB
r
=
=> r =  ----- (1)
mv
Bq
 
Time period = 
2 2 2
 =  = 
r rm m
v Bqr Bq
p p p
V = 
1
 => frequency is independent of velocity
2
Bq
T m p
=
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