NAB(-0 -NAB,;
Since we have assumed that B is increasing with
dt dt
time, dB /dt is a positive quantity. Also, A=|A |0s
25 x 1.25
=-208.3 V
positive by definition. Hence, the right hand side of 0.15
Eq. (1) is a negative quantity. (4
.. The magnitude of the induced emf.
The right hand rule for area vector fixes the
positive sense of circulation around the loop as the = 208.3 V
clockwise sense. Then, by Lenz's law the induced
current in the loop is in the anticlockwise sense. The Arectangular coil of length 0.5 mand
(2) breada
sense of the induced emf is the same as the sense of
0.4 mhas resistance of 5 N. The coil is placed
the current it drives. With the clockwise sense fixed ina
magnetic feldlof induction 0.05 T and its
as positive, the anticlockwise sense of the induced directioa
is perpendicular to the plane of the
current is negative. Hence, the sense of e is also
magnetic induction is uniformly reduced fo
negative. That is, the left hand side of Eq. (1) is
in 5 milliseconds, find the emf and curtend
indeed a negative quantity. Thus, introducing a induced in the coil. (3 marks)
minus sign in Faraday's second law incorporates
Solution:
Lenz's law into Faraday's law.
Data : l=0.5 m, b =0.4 m, R=52, B, =0.05T
*Q. 20. A uniform magnetic field B, pointing upward
B,=0, dt=5 x 10-s
flls a circular region of radius s in horizontal
plane. If B changes with time, ind the induced Area of the coil, A=lb=0.5 x 0.4=0.2 m²
emf. (1 mark) Initial magnetic flux, = AB,
Ans. The area of the region, A= 1s, remains constant =0.02 x 0.05 =0.01 W%
while B= B(t) is a function of time. Therefore, the Final magnetic flux, ;=0 (:: B=0)
induced emf, do
.:. The induced emf,e = dt -(,-4)
(BA) =-A dB(t) ,dB(+)
dm dt
dt dt di dt
(0-0.01)
5x 10-32V
Note : Emf and electric field are diferent physical
quantities, whose respective SI units are the volt and the 2
volt per metre. The question has accordingly been correc The induced current, I= R 5 0 4 A
ted.]
(3) Asquare wire loop with sides 0.5 mis placed with
Solved Problems 12.1-12.4
its plane perpendicular to a magnetic field. The
Q. 21. Solve the following: resistance of the loop is 52. Find at what rate the
(1) A coil of effective area 25 m² is placed in a magnetic induction should be changed so thali
field-free region. Subsequently, a uniform current of 0.1 A is induced in the loop. (2 marks)
magnetic field that rises uniformly from zero to Solution:
1.25 T in 0.15 s is applied perpendicular to the Data : l=0.5 m, R=5 2, I=0.1A
plane of the coil. What is the magnitude of the emf A=1'= 0.5 x 0.5 =0.25 m²
induced in the coil ? (2 marks ) The magnitude of the induced emf,
Solution: do d dB
lel = dt dt (BA) = A
Data : NA= 25 m, BË=1.25 T, B;=0, At=0.15s dt
Initial magnetic flux, =0 (:: B=0) since the area (A) of the coil is constant. Ih
Final magnetic flux, ;= NAB, lel A dB
induced current, I =
do (-4) R R dt
e= dt inducion
dt . The time rate of change of magnetic
272 NAVNEET PHYSICS DIGEST : STANDARD XII (PART II)
, AB IR 0.1 x 5
A 0.25
-2 T/#
dt
Induced emf, e =
The magnetic flux through a loop of
is varying according to the resistance
0,154 -0.1452 8.8 x 10-x 2
0.12
Q=6+ 7t+1, where
relation 0.5
is in milliweber and = 1.76 x 10-2V
t is
Insecond. What is the emf induced in the
loop at
(=la and the magnitude of the current? (3 (6) A 1000 turn, 20 cm diameter coil is rotated in th¹
marks)
solutlon : Earth's magnetic fleld of strength 5 x 105 T. The
Data: R
R=0.1
: 2, Om =62 +71+1 mWb, t =1s plane of the coilwas initially perpendicular to the
Earth's field and is rotated to be parallel to the
6) The induced emf, |e| dm
di (6 +7t +1) field in 10 ms? Find the average emf induced.
= (12 +7) mV (3 marks)
= 12(1) +7=19 mV Solution:
Data : N= 1000, d = 0.2 m, B=5x10-ST,
(i1) The magnitude of the current lel R
Af 10 ms = 102 s
19 mV
Radius of coil, r =d/2= 10 m
=190 mA
0.1Q Induced emf, e= -NN4
At At
0 A
wire 88 cm long ls bent lnso a circular loop and Initial area, A, = ny' and initial flux,
kept with its plane perpendicular to a magnetie NO, = NBA, = NB (u')
feld of induction 2,5 Wb/m², Within 0.85 second, Final Alux, =0, since the plane of the coil is
the coll is changed to a square and the magnetic parallel to the field lines.
Induction is inereased by 0.3 Wb/m', Calculate the 0- BA, NB
At
emf induced in the wire, (3 marks )
10-5)(3.l42 x 10-2) =
Bolutlon : =(10X5 x 10-2
0.1571 V
Data : =88 cm, B, = 2.5 Wb/m², B,=3 Wb/m²,
At=0.5 s
(7) A 8earch coil having 2000 turns with area 1.5 cm²
Por the circular loop, l= 2rr is placed in a magnetic field of 0.60 T. The coil is
88 moved rapidly out of the ield in a time of 0.2 s.
*22x(221714 cm =0.14 m
Calculate the induced emf in the search coil.
.. Area of the circular loop, A, = r (2 marks)
Solution:
-(0.14)²= 0.0616 m²
Data : N= 2000, A = 1.5 x 10-* m², A, = 0,
.. Initial magnetic flux, 4 = A,B; B=0.6T, Af = 0.2 s
=0.0616 x 2.5 =0,154 Wb Initial flux, Ng = NBA, = 2000(0.6)X1.5 x 10-*)
=0.18 Wb
For the square loop, length of each side
Final flux, NO = 0, since the coil is withdrawn
88 cm = 22 cm = 0.22 m
4 out of the field.
* Area of the square loop, A, = (0.22) Induced emt, e =-Nn
=0.0484 m² At At
0-0.18
=0,9 V
. Final magnetic flux, ; = AfB 0.2
0.0484 x3 =0,1432 Wb
E028 273
Narneet Physics Digest : Std. XII (Part I)
Since we have assumed that B is increasing with
dt dt
time, dB /dt is a positive quantity. Also, A=|A |0s
25 x 1.25
=-208.3 V
positive by definition. Hence, the right hand side of 0.15
Eq. (1) is a negative quantity. (4
.. The magnitude of the induced emf.
The right hand rule for area vector fixes the
positive sense of circulation around the loop as the = 208.3 V
clockwise sense. Then, by Lenz's law the induced
current in the loop is in the anticlockwise sense. The Arectangular coil of length 0.5 mand
(2) breada
sense of the induced emf is the same as the sense of
0.4 mhas resistance of 5 N. The coil is placed
the current it drives. With the clockwise sense fixed ina
magnetic feldlof induction 0.05 T and its
as positive, the anticlockwise sense of the induced directioa
is perpendicular to the plane of the
current is negative. Hence, the sense of e is also
magnetic induction is uniformly reduced fo
negative. That is, the left hand side of Eq. (1) is
in 5 milliseconds, find the emf and curtend
indeed a negative quantity. Thus, introducing a induced in the coil. (3 marks)
minus sign in Faraday's second law incorporates
Solution:
Lenz's law into Faraday's law.
Data : l=0.5 m, b =0.4 m, R=52, B, =0.05T
*Q. 20. A uniform magnetic field B, pointing upward
B,=0, dt=5 x 10-s
flls a circular region of radius s in horizontal
plane. If B changes with time, ind the induced Area of the coil, A=lb=0.5 x 0.4=0.2 m²
emf. (1 mark) Initial magnetic flux, = AB,
Ans. The area of the region, A= 1s, remains constant =0.02 x 0.05 =0.01 W%
while B= B(t) is a function of time. Therefore, the Final magnetic flux, ;=0 (:: B=0)
induced emf, do
.:. The induced emf,e = dt -(,-4)
(BA) =-A dB(t) ,dB(+)
dm dt
dt dt di dt
(0-0.01)
5x 10-32V
Note : Emf and electric field are diferent physical
quantities, whose respective SI units are the volt and the 2
volt per metre. The question has accordingly been correc The induced current, I= R 5 0 4 A
ted.]
(3) Asquare wire loop with sides 0.5 mis placed with
Solved Problems 12.1-12.4
its plane perpendicular to a magnetic field. The
Q. 21. Solve the following: resistance of the loop is 52. Find at what rate the
(1) A coil of effective area 25 m² is placed in a magnetic induction should be changed so thali
field-free region. Subsequently, a uniform current of 0.1 A is induced in the loop. (2 marks)
magnetic field that rises uniformly from zero to Solution:
1.25 T in 0.15 s is applied perpendicular to the Data : l=0.5 m, R=5 2, I=0.1A
plane of the coil. What is the magnitude of the emf A=1'= 0.5 x 0.5 =0.25 m²
induced in the coil ? (2 marks ) The magnitude of the induced emf,
Solution: do d dB
lel = dt dt (BA) = A
Data : NA= 25 m, BË=1.25 T, B;=0, At=0.15s dt
Initial magnetic flux, =0 (:: B=0) since the area (A) of the coil is constant. Ih
Final magnetic flux, ;= NAB, lel A dB
induced current, I =
do (-4) R R dt
e= dt inducion
dt . The time rate of change of magnetic
272 NAVNEET PHYSICS DIGEST : STANDARD XII (PART II)
, AB IR 0.1 x 5
A 0.25
-2 T/#
dt
Induced emf, e =
The magnetic flux through a loop of
is varying according to the resistance
0,154 -0.1452 8.8 x 10-x 2
0.12
Q=6+ 7t+1, where
relation 0.5
is in milliweber and = 1.76 x 10-2V
t is
Insecond. What is the emf induced in the
loop at
(=la and the magnitude of the current? (3 (6) A 1000 turn, 20 cm diameter coil is rotated in th¹
marks)
solutlon : Earth's magnetic fleld of strength 5 x 105 T. The
Data: R
R=0.1
: 2, Om =62 +71+1 mWb, t =1s plane of the coilwas initially perpendicular to the
Earth's field and is rotated to be parallel to the
6) The induced emf, |e| dm
di (6 +7t +1) field in 10 ms? Find the average emf induced.
= (12 +7) mV (3 marks)
= 12(1) +7=19 mV Solution:
Data : N= 1000, d = 0.2 m, B=5x10-ST,
(i1) The magnitude of the current lel R
Af 10 ms = 102 s
19 mV
Radius of coil, r =d/2= 10 m
=190 mA
0.1Q Induced emf, e= -NN4
At At
0 A
wire 88 cm long ls bent lnso a circular loop and Initial area, A, = ny' and initial flux,
kept with its plane perpendicular to a magnetie NO, = NBA, = NB (u')
feld of induction 2,5 Wb/m², Within 0.85 second, Final Alux, =0, since the plane of the coil is
the coll is changed to a square and the magnetic parallel to the field lines.
Induction is inereased by 0.3 Wb/m', Calculate the 0- BA, NB
At
emf induced in the wire, (3 marks )
10-5)(3.l42 x 10-2) =
Bolutlon : =(10X5 x 10-2
0.1571 V
Data : =88 cm, B, = 2.5 Wb/m², B,=3 Wb/m²,
At=0.5 s
(7) A 8earch coil having 2000 turns with area 1.5 cm²
Por the circular loop, l= 2rr is placed in a magnetic field of 0.60 T. The coil is
88 moved rapidly out of the ield in a time of 0.2 s.
*22x(221714 cm =0.14 m
Calculate the induced emf in the search coil.
.. Area of the circular loop, A, = r (2 marks)
Solution:
-(0.14)²= 0.0616 m²
Data : N= 2000, A = 1.5 x 10-* m², A, = 0,
.. Initial magnetic flux, 4 = A,B; B=0.6T, Af = 0.2 s
=0.0616 x 2.5 =0,154 Wb Initial flux, Ng = NBA, = 2000(0.6)X1.5 x 10-*)
=0.18 Wb
For the square loop, length of each side
Final flux, NO = 0, since the coil is withdrawn
88 cm = 22 cm = 0.22 m
4 out of the field.
* Area of the square loop, A, = (0.22) Induced emt, e =-Nn
=0.0484 m² At At
0-0.18
=0,9 V
. Final magnetic flux, ; = AfB 0.2
0.0484 x3 =0,1432 Wb
E028 273
Narneet Physics Digest : Std. XII (Part I)
