MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT1FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
1. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is
suspended in mid-air by a uniform horizontal magnetic field B (see figure). What is the
magnitude of the magnetic field?
•
Solution:
The upward force F of magnitude IlB must be balanced by the force due to gravity:
Mg = IlB
B = Mg/Il
=
The magnetic field is 0.65 T.
2. If the magnetic field is parallel to the positive y-axis and the charged particle is
moving along the positive x-axis as in figure. Which way would the Lorentz force be for (a)
an electron (b) proton
1
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:1 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT2FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
Solution:
The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v
is along the z-axis (right hand thumb rule). So, (a) for electron it will be along –z axis (b) for a
proton the force is along +z axis.
B
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
3. What is the radius of the path of an electron moving at a speed of 3 × 107 m/s in a
magnetic field of 6 × 10-4 T perpendicular to it? What is its frequency? Calculate its energy
in keV.
•
Solution:
We use the equation:
R = mv/(qB) = 9 × 10-31 kg × 3 × 107 ms-1 /(1.609 × 10-19 C × 6 ×10-4 T)
= 26 × 10-2 m =26 cm
ν = v/(2πr) = 2 × 106 s-1 = 2 × 106 MHz
E = (1/2) mv2 = (1/2) 9 10-31 kg × 9 × 1014 m2/s2 = 40.5 × 10-17 J
= 4 × 10-16 J = 2.5 keV.
4. A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic
field for accelerating protons? If the radius of its ‘dees’ is 60 cm, what is the kinetic energy
(in MeV) of the proton beam produced by the accelerator.
•
Solution:
The oscillator frequency should be same as proton's cyclotron frequency, and we have
B=2
Final velocity of protons is
V=r
2
E = (1/2) mv2 = 1.67 × 10-27 × 14.3 × 1014 / (2 × 1.6 × 10-13) = 7 MeV.
5. An element ∆l = ∆x is placed at the origin and carries a large current I = 10 A as in
figure. What is the magnetic field on the y-axis at a distance of 0.5 m. ∆x = 1 cm
2
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:2 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT3FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
Solution:
.
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
The direction of the field is in the +z direction. This is so since,
.
6. A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0
cm as shown in the figure. Consider the magnetic field B at the center of the arc. (A) What
is the magnetic field due to the straight segments? (b) In what way the contribution to B
from the semicircle differs from that of a circular loop and in what way does it resemble?
(c) Would your answer be different if the wire were bent into a semi-circular arc of the
same radius but in the opposite way as shown in the figure?
•
Solution:
(a) dl and r for each element of the straight segments are parallel. Therefore, dl
segments do not contribute to l B l.
r = 0. Straight
(b) For all segments of the semicircular arc, dl
r are all parallel to each other. All such
contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the righthand rule and magnitude is half that of a circular loop. Thus B is 1.9
10-4 T normal to the plane of
the paper going into it.
(c) Same magnitude of B but opposite in direction to that in (b).
in
7. Consider a tightly wound 100 turn coil of radius 10 cm carrying a current of 1 A. What
is the magnitude of the magnetic field at the center of the coil?
•
Solution:
Since the coil is tightly wound, we take each circular element to have the same radius R = 10 cm =
0.1 m. The number of turns N = 100. The magnitude of the magnetic field is,
B=
.
3
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:3 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT4FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
8. The figure shows a long straight wire of a circular cross-section (radius a) carrying
steady current I. The current I is uniformly distributed across this cross-section. Calculate
the magnetic field in the region r<a and r>a.
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
•
Solution:
(a) Consider the case r < a. The Amperian loop, labeled 2, is a circle concentric with the crosssection. For this loop,
L=2 r
Ie = Current enclosed by the loop = I
The result is the familiar example for a long straight wire
B(2
r) =
B=
B
(r>a)
(b) Consider the case r<a. The Amperian loop is a circle labeled 1. For this loop, taking the radius of
the circle to be r,
L=2 r
Now the current enclosed Ie is not I, but less than this value. Since the current distribution is
uniform, the current enclosed is,
Ie = I
Using Ampere's law,
B(2
r) =
B=
B
(r<a)
The plot of the magnitude of B with distance r from the center of the wire increases linearly upto 'a'
and then decreases exponentially. The direction of the field is tangential to the respective circular
loop (1 or 2) and given by the right-hand rule.
9. A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries
a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?
4
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:4 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT5FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
Solution:
The number of turns per unit length is,
N=500/0.5 = 1000 turns / m.
The length l = 0.5 m and radius r = 0.01 m. Thus, l/a = 50 i.e., l>>a. Hence, we can use the long
solenoid formula:
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
B=
=
= 6.28
.
10. The horizontal component of the earth’s magnetic field at a certain place is 3.0
105 T and the direction of the field is from the geographic south to the geographic north. A
very long straight conductor is carrying a steady current of 1 A. What is the force per unit
length on it when it is placed on a horizontal table and the direction of the current is (a)
east to west (b) south to north?
•
Solution:
F = Il
B
F = Il B sin θ
The force per unit length is
f = F / l = IB sin θ
(a)When the current is flowing from east to west, θ = 90ο
Hence, f = IB = 1
3
10-5 = 3
10-5 Nm-1
This is larger than the value 2 ´ 10-7 Nm-1 quoted in the definition of Ampere. Hence it is important
to eliminate the effect of the earth’s magnetic field and other stray fields while standardizing the
ampere.
(b)When the current is flowing from south to north, θ = 0ο, f = 0.
Hence there is no force on the conductor.
11. A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A
(a) What is the field at the center of the coil?
(b) What is the magnetic moment of this coil?
The coil is placed in a vertical plane and is free to rotate about a horizontal axis which
coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists
such that initially the axis of the coil is in the direction of the field. The coil rotates through
an angle of 90οunder the influence of the magnetic field.
(c) What are the magnitudes of the torques on the coil in the initial and final position?
(d) What is the angular speed acquired by the coil when it has rotated by 90ο? The moment
of inertia of the coil is 0.1 kg m2.
•
Solution:
(a) The magnetic field is given by,
B=
5
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:5 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT6FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
The direction is given by the right hand thumb rule.
(b) The magnetic moment is given by M = NIA = NI
=100
3.2 3.14
10-2 = 10 Am2
(c)
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
Initially, θ = 0. Thus, initial torque
(d) From Newton’s second law,
where I is the moment of inertia of the coil.Thus,
Integrating the above equation from 0 to
.
12. (a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform
magnetic field be set up in such a manner that the loop turns around itself.
(b) A current-carrying circular loop is located in a uniform external magnetic field. If the
loop is free to turn, what is its orientation, the flux of the total field is maximum.
(c) A loop of irregular shape carrying current is located in an external magnetic field. If the
wire is flexible, why does it change to a circular shape?
•
Solution:
(a) No, because that would require
to be in the vertical direction. But
=IA
B, and since A of
the horizontal loop is in the vertical direction,
would be in the plane of the loop for any B.
(b) Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of
external field. In this orientation, the magnetic field produced by the loop is in the same direction as
external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total
field.
(c) It assumes circular shape with its plane normal to the field to maximize flux, since for a given
perimeter, a circle encloses greater area than any other shape.
6
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:6 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT7FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
13. In the circuit given, the current is to be measured. What is the value of the current if
the ammeter shown (a) is a galvanometer with a resistance RG = 60.00 Ω (b) is a
galvanometer described in (a) but converted to an ammeter by a shunt resistance rs = 0.02
Ω (c) is an ideal ammeter with zero resistance?
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
•
Solution:
(a) Total resistance in the circuit is
RG+3 = 63 Ω. Hence, I = 3/63 = 0.048 A
(b) Resistance of the galvanometer converted to an ammeter is,
Total resistance in the circuit is,
0.02 Ω+3 Ω = 3.02 Ω. Hence, I = 3/3.02 = 0.99 A.
(c) For the ideal ammeter with zero resistance,
I = 3/3 = 1.00 A.
14. A circular coil of wire consisting of 100 turn-s, each of radius 8.0 x 10-2 m carries a
current of 0.40 A.What is the magnitude of the magnetic field B at the centre of the coil?
•
Solution:
Number of turns 'N' = 100
Radius of the coil 'r' = 8.0 × 10-2 m
Current in the coil 'I' = 0.40 A
... The magnitude of the magnetic field 'B' is given by
= 3.1 x 10-4 Tesla.
15. A long straight wire carries a current of 35A. What is the magnitude of the field B at
a point 0.20 m from the wire?
7
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:7 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT8FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
Solution:
Current through the wire I = 35A
Distance of separation 'r' = 0.20 m
... The magnitude of the field
= 3.5 x 10-5 T.
.
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
16. A long straight wire in the horizontal plane carries a current of 50 A in north to south
direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
•
Solution:
Current through the wire I = 50A
Distance of separation 'r' = 2.5 m
...The magnitude of the field =
The direction of B is vertically upwards.
= 4 x 10-6 T
17. A horizontal overhead power line carries a current of 90 A in an east to west
direction. What is the magnitude and direction of the magnetic field, due to the current, 1.5
m below the line?
•
Solution:
The current I = 90 A
Distance below the power line 'r' = 1.5 m
... The magnitude and direction of the magnetic field is given by
= 1.2 x 10-5 T
Applying the right-hand thumb rule, we find that the magnetic field at the observation point is
directed towards south.
18. What is the magnitude of the magnetic force per unit length on a wire carrying a
current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of
0.15T?
•
= BI sin q
= 0.15 × 8 ×€sin 30°
= 4 × 0.15 Nm-1
= 0.6 Nm-1.
in
Solution:
Current in the wire 'I' = 8 A
The angle 'q' = 30°
Magnitude of the magnetic field = 0.15 T
... The magnitude of the magnetic force per unit length on the wire is given by
19. A 3.0 x 10-2 m wire carrying a current of 10 A is placed inside a solenoid
perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What
is the magnetic force on the wire?
8
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:8 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT9FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
Solution:
Length of the wire 'l' = 3.0 × 10-2 m
Current in the wire 'I' = 10 A
The magnetic field 'B' = 0.27 T
... The magnetic force on the wire 'F' = BIl
= 0.27 × 10 × 3.0 × 10-2 N
= 8.1 × 10-2 N
The direction of force is obtained by applying by Fleming's Left Hand Rule.
20. Two long and parallel straight wires A and B carrying currents of 8.0A and 5.0A in
the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm
section of wire A.
•
Solution:
Current through the wire A 'I1' = 8.0A
Current through the wire B 'I2' = 5.0A
Separation between the two wires 'r' = 4.0 cm = 0.04 m
Force per unit length 'F' =
= 2 × 10-4 Nm-1
Force on 10 cm section of wire A = 2 × 10-4 × 0.10 N
= 2 × 10-5 N
Since the currents are in same direction, the force is attractive; its direction is normal to A and B.
21. A closely wound solenoid 0.80 m long has 5 layers of windings of 400 turns each. The
diameter of the solenoid is 1.8 x 10-2 m. If the current carried is 8.0 A, estimate the
magnitude of magnetic field inside the solenoid near its centre.
•
Solution:
Number of turns 'N' = 400 × 5 = 2000
Length of the solenoid 'l' = 0.80 m
Diameter of the solenoid = 1.8 × 10-2 m
Current in the solenoid 'I' = 8.0 A
... The magnitude of the magnetic field inside the solenoid near its centre is given by
B = m0nI =
= 2.5 x 10-2 T.
22. A square coil of side 10 cm consists of 20 turns and carries a current of 12A. The coil
is suspended vertically and the normal to the plane of the coil makes an angle of 30° with
the direction of a uniform horizontal magnetic fields of magnitude 0.80T. What is the
magnitude of torque experienced by the coil?
•
Solution:
Number of turns 'N' = 20
Current through the square coil 'I' = 12A
Magnitude of the magnetic field 'B' = 0.80 T
Angle between normal to plane of coil and the magnetic field 'q' = 30°
Side of the square coil 'a' = 10 cm
= 0.10m
Area of the coil 'A' = a2
= (0.10)2
= 0.01 m2
. The magnitude of the torque is given as
9
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:9 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT10
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
t = NBIA sin q
= 20 × 0.80 ×€12 × 0.01 × sin 30°
= 0.96 Nm.
23. Two moving coil meters M1 and M2 have the following particulars:
R1 = 10 W, N1 = 30,
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
A1 = 3.6 × 10-3 m2, B1 = 0.25 T
R2 = 14 W, N2 = 42,
A2 = 1.8 ×€10
×€ -3 m2, B2 = 0.50 T,
(The spring constants are identical for the two meters). Determine the ratio of
(a) current sensitivity and
(b) voltage sensitivity of M2 and M1.
•
Solution:
(a) Current sensitivity of a meter
=
(Since k1 = k2 )
= 0.714
Therefore the ratio of current sensitivities of M2 : M1 = 1/ 0.714 = 1.4
(b) Voltage sensitivity of a meter
=
=
= 0.714 × (14/10)
= 0.9996.
Therefore the ratio of the voltage sensitivities of M2 : M1 = 1/0.9996 =1.
24. In a chamber a uniform magnetic field of 6.5G (1G = 10-4 T) is maintained. An
electron is shot into the field with a speed of 4.8 x 106ms-1 normal to the field. Explain why
the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 1019
C, me = 9.1×
× 10-31 kg).
10
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:10 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT11
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
•
Solution:
The path of the electron is a circle because the given field is magnetic and the electron is shot
normal to the field.
Strength of the magnetic field 'B' = 6.5 G
= 6.5 × 10-4 T
Speed of the electron shot into the field 'u' = 4.8 × 106 ms-1
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
Radius of the circular path of the electron 'r'=
For electron, me = 9.1 × 10-31 kg
e = 1.6 × 10-19 C
... r =
m
= 4.2 × 10-2 m
Therefore the radius of the circular orbit 'r' = 4.2 × 10-2.
25. For problem number 24, obtain the frequency of revolution of the electron in its
circular orbit. Does the answer depend on the speed of the electron? Explain.
•
Solution:
Frequency of revolution =
= 18 MHz
It does not depends on the speed of the electron, because the radius of the circular path increases in
direct proportion to the velocity of the electron. As a result, time period of the electron and hence its
frequency remains independent of the speed of the electron.
26. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended
vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an
angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that
must be applied to prevent the coil from turning.
(b) Would your answer change if the circular coil in (a) were replaced by a planar coil of
some irregular shape that encloses the same area?
•
Solution:
(a) Number of turns 'N' = 30
Radius of the circular coil 'r' = 8.0 cm
= 0.08 m
Current through the coil 'I' = 6.0A
Magnitude of the magnetic field 'B' = 1.0 T
Angle between normal to plane of coil and magnetic field, θ = 60°
Area of the coil, A = πr2
= 3.14 (0.08)2 m2
= 0.02 m2
.
. . The magnitude of the torque t = NBIA sin θ
= 30 × 3.14 × 64 × 10-4 × 0.866 Nm
= 3.133 Nm
(b) No, the answer remains unchanged because the formula t = NIA × B is independent of the shape
and remains constant as long as the area does not change.
11
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:11 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT12
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
27. Two concentric circular coils X and Y of radii 16 cm and 10 cm respectively lie in the
same vertical plane containing the north-south direction. Coil X has 20 turns and carries a
current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of current in X
is anticlockwise and in Y clockwise, for an observer looking at the coils facing west. Give
the magnitude and direction of the net magnetic field due to the coils at their center.
•
Solution:
.
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
Let By be the magnetic field at the centre of the coil, due to the current Iy flowing in it. If Ny be the
number of turns and ry the radius of the coil; then
By =
tesla
But m0 = 4p × 10-7 SI units
Ny = 25
Iy = 18 A
ry = 0.1 m
... B =
= 9p x 10-4 tesla
For an observer looking at the coil Y facing west, the current in coil y is flowing in the clockwise
direction. So, the direction of By is towards west.
If Bx is the magnetic field at the centre of coil x, then
= 4p × 10-4 T
in
Bx =
For an observer at the coil X facing west, the current in coil x is flowing in the anticlockwise direction.
So, the direction of Bx is towards east.
Bx and By act along the same line but in the opposite directions. So, the resultant magnetic
field B will be in the direction of the bigger of the two vectors. The magnitude of B i.e. B will be equal
to the difference of the magnitudes of Bx and By
... B = By - Bx = (9p × 10-4) - (4p × 10-4)
= 5p × 10-4 tesla
12
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:12 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT13
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
= 1.57 × 10-3 tesla
will be directed towards the west.
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
28. A magnitude field of 100 G (1G = 10-4 T) is required which is uniform in a region of
linear dimension about 10 cm and area of cross-section about 10-3 m2. The maximum
current-carrying capacity of a given coil of wire is 15 A and the number of turns m-1.
Suggest some appropriate design particulars of a solenoid for the required purpose.
Assume the core is not ferromagnetic.
•
Solution:
A solenoid with length of 50 cm, radius about 4 cm, number of turns about 400 and current about 10
A is required. Such an arrangement will be appropriate for the purpose the solenoid will serve .
(The above said particulars are appropriate).
29. For a circular coil of radius R and N turns carrying current I, the magnitude of the
magnetic field at a point on its axis at a distance x from its center is given by
(a) Show that this reduces to the familiar result for field at the center of the coil.
(b) Consider two parallel co-axial circular coils of equal currents in the same direction, and
separated by a distance R. Show that the field on the axis around the mid-point between
the coils is uniform over a distance that is small compared to R, and is given by
B = 0.72
•
approximately.
Solution:
(a) At the center of the coil, x = 0
... B =
B=
(b) In a small region of length 2d, about the mid-point between the coils,
13
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:13 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT14
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
(The terms containing d2/R2 and higher powers of d/R are neglected since d/R << 1)
1) The terms linear in d/R cancel giving a uniform field B in a small region:
B=
B = 0.72
30. A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26
cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is
the magnetic field
(i) outside the toroid,
(ii) inside the toroid, and
(iii) in the empty space surrounded by the toroid.
•
Solution:
Inner radius of the toroid = 25 cm = 0.25 m
Outer radius of the toroid = 26 cm = 0.26 m
... Mean radius of the toroid 'r' =
Total number of turns 'N' = 3500
Current in the wire 'I' = 11 A
Number of turns per unit length 'n' =
m = 25.5 x 10-2 m
= 2185 (approximately)
(a) the magnetic field outside the toroid is zero.
(b) Magnetic field inside the core of the toroid,
B = m0 nI
= 4π × 10-7 × 2185 × 11 tesla
= 3 × 10-2 T
(c) The field in the empty space surrounded by the toroid is zero.
14
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:14 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT15
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
31. (a) A magnetic field that varies in magnitude from point to point but has a constant
direction (east to west) is set up in a chamber. A charged particle enters the chamber and
travels undeflected along a straight path with constant speed. What can you say about the
initial velocity of the particle?
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
(b) A charged particle enters an environment of a strong and non-uniform magnetic field
varying from point to point both in magnitude and direction, and comes out of it following a
complicated trajectory. Would its final speed equal the initial speed if it suffered no
collisions with the environment?
(c) An electron-traveling west to east enters a chamber having a uniform electrostatic field
in north to south direction. Specify the direction in which a uniform magnetic field should
be set up to prevent the electron from deflecting from its straight line path.
•
Solution:
(a) If it travels undeflected the charged particle must be moving in the direction of the field. Thus,
the initial velocity must be either parallel or anti parallel to magnetic field.
(b) If the charged particle suffered no collisions with the environment then the final speed equals the
initial speed, because magnetic force can change the direction of the speed, not its magnitude.
(c) Magnetic field should be in a vertically downward direction to prevent the electron from deflecting
from its straight-line path.
32. An electron emitted by a heated cathode and accelerated through a potential
difference of 2.0 kV enters a region with uniform magnetic field of 0.15 T. Determine the
trajectory of the electron if the field is
(a) transverse to its initial velocity
(b) makes an angle of 30° with the initial velocity. Take mass of electron = 9 × 10-31 kg.
•
Solution:
Potential difference 'V' = 2.0 kV
= 2000 volt
Strength of magnetic field 'B'= 0.15T
We know that
= eV
u2 =
= ms-1
= (8/3) × 107 ms-1
(a) When the magnetic field is transverse to the initial velocity
The magnetic force Beu shall provide the necessary centripetal force for the circular motion of the
electron.
Beu =
or r =
15
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:15 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT16
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
r=
m
= 1 mm
Thus the electron travels in a circular path of radius 1mm normal to B.
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
(b) When the magnetic field makes an angle of 30° with the initial velocity
The component of velocity perpendicular to the field shall make the electron move along a circle. On
the other hand, no forces shall act on the electron due to component of velocity in the direction of
the field. So, the electron shall follow a straight path. The combined effect will be that the electrons
shall follow a straight path. The combined effect will be that the electron shall follow a helical path.
Radius of helical path, r =
=
= 1 × (1/2)mm
= 0.5 mm.
33. A magnetic field set up using Helmholtz coils is uniform in a small region and has a
magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a
direction normal to the common axis of the coils. A narrow beam of (single-species)
charged particles all accelerated through 15 kV enters this region in a direction
perpendicular to both the axis of the coils and the electrostatic field. If the beam remains
undeflected when the electrostatic field is 9.0 × 105 Vm-1, make a simple guess as to what
the beam contains. Why is the answer not unique?
•
Solution:
Magnitude of the magnetic field 'B' = 0.75 T
Electrostatic field 'E' = 9.0 × 105 V m-1
Accelerated energy of the charged particles = 15 keV
i.e. (1/2)mu2 = 15 keV
= 15 × 103 × 1.6 10-19 J
= 24 × 10-16 J
Since the given beam remains undeflected,
Beu = Ee
u=
=
= 1.2 × 106 ms-1
.
. . (1/2)m (1.2 × 106)2 = 24 × 10-16
m=
= 3.33 × 10-27 kg
The mass of the given charged particle is clearly double the mass of the proton. If the charge on the
given particle is 1.6 ×€10-19 C, then the given charged particles may be deutrons. However, the
16
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:16 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT17
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
answer is not unique. This is because He++
and Li+++
also have the same value of
.
34. A straight horizontal conducting rod of length 0.45 m and mass 0.06 kg is suspended
by two vertical wires at its ends. A current of 5.0 A set up in the rod through the wires.
.
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
(a) What magnetic field should be set up normal to the conductor in order that the tension
in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed, keeping
the magnetic field same as before? Neglect the mass of the wires.
•
Solution:
Length of the rod 'l' = 0.45 m
Current through the conductor 'I' = 5.0 A
Mass of the rod 'm' = 0.06 kg
Acceleration due to gravity 'g' = 9.8 N
(a) Force required to balance the weight of the rod,
F = m × g = 0.06 × 9.8 N = 0.588 N
We know that F = BIl
or B =
= 0.26 T
This much magnetic field should be applied in such a way that the force due to the magnetic field
acts upwards on the rod.
(b) Keeping the magnetic field same as in part (a), if the direction of current is reversed, then the
force due to the magnetic field acts downwards.
Now, total tension in the wires = Force due to weight of rod + force due to magnetic field
= 0.588 N + 0.588 N
= 1.176 N.
35. The wires which connect the battery of an automobile to its starting motor carry a
current of 300 A(for a short time). What is the force per unit length between the wires if
they are 0.7m long and 0.015 m apart? Is the force attractive or repulsive? Also determine
the force on each wire.
•
We know that Force per unit length F =
in
Solution:
Current through the wires I1 = I2 = 300 A
Length of the wire l = 0.70 m
Separation between the two wires 'r' = 0.015 m
=
= 1.2 Nm-1
... Force on each wire = 0.7 × 1.2 N = 0.84 N.
The value 0.84N is approximate. Since the formula holds strictly for infinitely long wires, so the
currents in the two wires are in opposite direction and hence the force is repulsive.
17
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:17 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT18
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
36. A uniform magnetic field of 3000 G is established along the positive z-direction. A
rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on
the loop in the different cases shown in the figure? What is the force on each case? Which
case corresponds to stable equilibrium?
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
•
Solution:
Strength of the magnetic field 'B' = 3000 Gauss = 3000 × 10-4 T
Area of the Rectangular loop 'A' = 10 × 5 cm2 = 50 × 10-4 m2
Current in the rectangular loop = 12 A
Angle between the plane of loop and direction of magnetic field 'θ' = 0°
(a) Torque on the loop 'τ' = BIA cos θ
= 3000 × 10-4 × 12 × cos 0°
= 1.8 × 10-2 N m
Now it is directed along y-axis.
(b) 1.8 × 10-2 Nm along y-direction.
(c) The torque τ = 1.8 × 10-2 N m along x-direction
(d) Torque 'τ' = 1.8 × 10-2 N m at an angle of 240° with positive x-direction.
(e) τ is zero ( Angle between plane of loop and direction of magnetic field is 90°)
(f) Zero, as all the four arms are perpendicular to the field.
Force in the different cases will be as follows:
(a) F = I (l × B) = IlB sin 90o.
(b), (c), (d), (e), (f) is zero
Only case (e) corresponds to stable equilibrium, while case (f) corresponds to unstable equilibrium.
37. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of
0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
18
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:18 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT19
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field. (The coil is made of
copper wire of cross-sectional area 10-5 m2, and the free electron density in copper is given
to be about 1029 m-3).
.in
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
•
Solution:
Number of turns in the coil 'N' = 20
Radius of the coil 'r' = 10 cm = 0.10 m
Area of the coil 'A' = πr2 = 3.14 × (0.10)2 = 0.031 m2
Strength of the magnetic field 'B' = 0.10 T
Current in the coil 'I' = 5.0 A
(a) The total torque on the coil 't' = NBIA cos 90° = 20 × 0.10 × 5.0 × 0.031 × 0 = 0.
(b) The net force on a planar current loop in a uniform magnetic field is always zero.
(c) Force on each electron is given by
F = Beu =
=
=
This force is the magnetic force alone.
N = 5 × 10-25 N
38. A solenoid 60 × 10-2 m long and of radius 0.04 m has 3 layers of windings of 300
turns each. A 2 × 10-2 m long wire of mass 2.5 × 10-3kg lies inside the solenoid near its
center normal to its axis. Both the wire and the axis of the solenoid are in the horizontal
plane. The wire is connected through two leads parallel to the axis of the solenoid to an
external battery which supplies a current (with appropriate sense of circulation) in the
windings of the solenoid can support the weight of the wire? Given g = 9.8 ms-2.
•
Solution:
Let I be the current required to be passed through the solenoid. The magnetic field set up at the
center of the solenoid is given by
B = µ0nI
This field acts normally to the length of the current-carrying wire. As a result, the wire experiences a
force F given by
F = BIl
where I′ is the current flowing through the wire and l is the length of the wire. The direction of
current through the solenoid and hence the magnetic field are in such a direction that the force F
acts upwards. The wire will be supported only if the force F is equal to the weight mg of the wire.
BI′ l = mg
or m0nII′ l = mg
or I =
where, m = 2.5 × 10-3 kg
g = 9.8 ms-1
µ0= 4π × 10-7 T m A-1
19
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:19 rktctb@yahoo.com web site www.mathematic.in
MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT20
FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
n=
turns m-1 = 1500 turns m-1
I′ = 6.0 A
l = 2.0 cm = 0.02 m
= 108.3 A.
in
.
ic
at
m
he 5
at 31
.m 44
w 1
w 0
w 81
09
... I =
20
CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in south and west delhi
EMAIL:20 rktctb@yahoo.com web site www.mathematic.in