CBSE Class 12 Physics HOTs Magnetic Effects Of Current and Magnetism

UNIT -3

MAGNETIC EFFECTS OF CURRENT & MAGNETISM

ONE MARK QUESTIONS

Question. An electric current of 0.25 A flows in a loop of radius 0.2 cm. Calculate the magnitude of the magnetic dipole moment of the dipole formed.
Answer : 3.14 x 10^-6 Am2 

Question. A circular coil of 'N' turns and diameter 'd' carries a current 'I'. It is unwound and rewound to make another coil of diameter '2d', current 'I' remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil.
Answer : Magnetic moment (m) = NIA , mB/mA = 2/1

Question. What will be the path of a charged particle moving in a uniform magnetic field at any arbitrary acute angle?
Answer : Helical 

Question. An electron and a proton, moving parallel to each other in the same direction with equal
momenta, enter into a uniform magnetic field (perpendicular to the plane of the paper)which is at right angles to their velocities. Trace their trajectories in the magnetic field.
Answer : Both proton and electron will make circular path of equal radii in anticlockwise as well as clockwise direction respectively.

Question. Why electrons can’t be accelerated by using a cyclotron?
Answer : Because relativistic mass of electrons become very high due to its very high velocity , which violates the condition of magnetic resonance.

Question. What is the function of the radial magnetic field in the moving coil galvanometer?
Answer : The arrangement provides linear scale to the galvanometer, because the plane of the coil always gets aligned parallel to the direction of the applied magnetic field.

Question. Two parallel wires carrying currents in the same direction attract each other, while the two beam of electrons travelling in the same direction repel each other. Give reason.
Answer : In case of parallel wires, only attractive magnetic interaction acts. In case of electron beams, the electrostatic repulsion is stronger than the attractive magnetic interaction.

Question. A given galvanometer is to be converted into (i) an ammeter (ii)a milliammeter (iii) a voltmeter.
In which case will the required resistance be (i)least (ii)highest
Answer : The required resistance has least value in case of an ammeter and maximum value in case of voltmeter.

Question. Why phosphor bronze alloy is preferred for the suspension wire of moving coil galvanometer?
Answer : Current sensitivity is more as it is given by k n BA I; Because k is small for phosphor bronze.

Question. What types of force is acting between two parallel wires carrying current in the same direction?
Answer : Force of attraction will act between these two current carrying wires.

TWO MARKS QUESTIONS:

Question. A solenoid of length 50 cm having 100 turns carrying a current of 4.5 A find the magnetic field (i) in the Interior of solenoid (ii) outside the solenoid.
Answer : (i) B = µ0ni = 6.28 x 10- 4 T (ii) B is almost zero.

Question. How will the magnetic field intensity at the centre of a circular coil carrying current change if the current through the coil is doubled and the radius of the coil is halved.
Answer : B = µ0n x 2I / 2 x (R/2) = 4B

Question. State Ampere’s circuital law.
Answer : It states that the line integral of magnetic field around any closed path in vacuum/air is µ0 times the total current threading the closed path.
Mathematically , we can express as

Question. A milliammeter whose resistance is 120 has full scale deflection with a current of 5mA . How will you use it to measure a maximum current of 100 A?
Answer : S = 0.006 Ω in parallel .

Question. A solenoid is 1m long and 3 cm in mean diameter. It has 5 layers of windings of 800 turns each and carries a current of 5 A. Find Magnetic Field Induction at the center of the solenoid.
Answer : 2.5 x 10 -2 T, parallel to the axis of the solenoid .

Question. Define current sensitivity and voltage sensitivity of a galvanometer.
Answer : Current sensitivity : It is defined as the deflection of coil per unit current flowing in it. Current
Sensitivity, Sθ=θ/I=(NAB )/C . Voltage sensitivity : It is defined as the deflection of coil per unit
potential difference across its ends.Voltage Sensitivty, SV=θ/V=NAB/GC ,where G is the resistance of galvanometer.

Question. What is the working principle of a moving coil galvanometer ?
Answer : The current carrying coil kept in a magnetic field always experiences a torque

Question Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle.
Answer : Expression for Period of Revolution : (Image 2

Question State Biot – Savart law and express this law in the vector form.
Answer :  dB = (µ0/ 4π) (I dl x r)/r

Question. Write the expression, in a vector form, for the magnetic Lorentz force F experienced by a charge q moving with velocity v in a magnetic field B. What is the direction of the magnetic force?
Answer :. F= q (VX)

THREE MARKS QUESTIONS

Question. A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120rev/min in a plane normal to the horizontal component of the earth’s magnetic field. The earth’s magnetic field at the place is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the 3txle and the rim of wheel. How will the value of the emf be affected if the number of spokes were increased?
Answer : (x) ANS: Horizontal component,
H = B cos θ
= 0.4 cos 60°
= 0.4 * ½
= 0.2 G
H = 0.2 * 10⁻⁴ T
The wheel is rotating in a plane normal to the horizontal component, so it will cut the horizontal component only, Vertical component of earth will contribute nothing in emf.
Thus, the emf induced is given as
e = ½ H ω L²,
Where ω = 2πN/t and
L = length of the spoke
=50 cm =0.5 m
E = ½ * 0.2 * 10⁻⁴ * (0.5)² * 2 *3.14 * 120 /60
E = 3.14 *10⁻⁵V
The value of emf induced is independent of the number of spokes as the emf’s across the spokes are in parallel. So, the emf will be unaffected with the increase in spokes.

Question. Three identical specimens of a magnetic materials nickel, antimony and aluminium are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.
Answer : The modifications are shown in the figure.
It happens because
(1) Nickel is a ferromagnetic substance.
(2) Antimony is a diamagnetic substance
(3) Aluminium is a paramagnetic substance.

Question. (1) What happens when a diamagnetic substance is placed in a varying magnetic field?
(2)Which of the substances are diamagnetic?
Bi, Al, Na, Cu, Ca and Ni
Answer : (1) When diamagnetic substance is placed in a varying magnetic field, it tends to move from stronger magnetic field to weaker magnetic field
(2) Bi and Cu.

Question. (1) How does angle of dip change as line goes from magnetic pole to magnetic equator of the earth?
(2) A uniform magnetic field gets modified as shown in fig .when two specimens X and Y are placed in it. Identify whether specimens X and Y are diamagnetic, paramagnetic of ferromagnetic.
Answer : (1) The angle of dip decreases from 90° to 0°.
(2) For paramagnetic material, no magnetic lines of force enter in it. So the specimen X is paramagnetic. For ferromagnetic materials, all magnetic lines of force prefer to go through it. So specimen Y is ferromagnetic.

Question. When two materials are placed in an external magnetic field, the behaviour of magnetic field lines is as shown in the fig. Identify the magnetic nature of these two materials.
Answer : (1)Material X is paramagnetic substance. when a specimen of a paramagnetic substance is placed in a magnetizing field, the lines of force prefer to pass through the specimen rather than through air. Thus magnetic induction inside the sample is more than the magnetic intensity.
(2)Material Y is a ferromagnetic substance. These are the substance in which a strong magnetism is reduced in the same direction as the applied magnetic field, these are strongly attracted by a magnet.

Question. Name the three elements required to specify the earth’s magnetic field at a given place. Draw a labeled diagram to define these elements. Explain briefly how these elements are determining to find out the magnetic field at a given place on the surface of the earth.
Answer : These are three elements determining earth’s magnetic field at any point of the earth.
a. Magnetic declination.
b. Magnetic dip.
c. Horizontal component of earth’s magnetic field.
Angle between geographical meridian and the magnetic meridian at any point is known as the magnetic declination at that point. Magnetic dip is the angle between the direction of earth’s magnetic field and the horizontal direction along the magnetic meridian.

Question. If χm stands for the magnetic susceptibility of a given material, identify the class of materials for which
a. -1 < χm < 0
b.0 < χm < Ɛ (where Ɛ stands for a small positive number)
(I) Write the range of relative magnetic permeability of the materials.
(II) Draw the pattern of magnetic field lines when the materials are placed in external magnetic field.
Answer : 38. a. -1 < χm < 0 = Diamagnetic material
0 < χm < Ɛ =Paramagnetic substance,
Relative permeability of diamagnetic material
μ = B/H where, 0 < μ < 1
For paramagnetic substance ,
μ> 1
but μ is not very large

Question. Name and define the elements of the earth’s magnetic field other than the horizontal component of the earth’s magnetic field. Why do we say that at the place like Mumbai and Delhi, a magnetic needle shows the true north direction quite accurately as compared to other places in India.
Answer : The two elements of earth’s magnetism other than horizontal components of magnetic field are
a. Magnetic Declination: The angle between the magnetic meridian and geographical meridian is known as the angle of declination at a given place of earth.
b. Angle of dip: The angle made by resultant of the earth’s magnetic field with the horizontal in magnetic meridian is known as angle of dip at the given place of earth.
The angle of dip is 90° at poles and 0°4 NE at Delhi and 0.58° NW at Mumbai. Thus, at both of these laces a magnetic needle shows the true north quite accurately.

FIVE MARKS QUESTIONS

Question. Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles. 
Answer : (a) If particle is performing circular motion due to magnetic force then centripetal force = Magnetic force
mv2/r = q ν B sin 90° ⇒ r = mν/qB ∴Time period = 2πr/v T = 2πm/qB
T = qB m2 π ∝ ν° ∴Frequency f = T 1 = m2 qB π ∝ ν°
(b) Cyclotron is a device to accelerate ions to extremely high velocities, by accelerating them repeatedly through high voltages. (Image 8)
Principle – A positive ion can acquire sufficiently large energy with a small alternating potential difference by making the ion cross the same electric field time and again by making use of a strong magnetic field.
Construction- It consists of a pair of hollow metal cylindrical chambers shaped like D, and called the Dees ; Both the Dees are placed under the circular pole pieces of a very strong electromagnet. The two Dees are connected to the terminals of a very high frequency and high voltage oscillator whose frequency is of the order of a few million cycles per second.
Working - Charged ion is passed through electric field again & again to be energised. Inside dee's strong ⊥ magnetic field turns the particle towards gap. So radius of semi circular path increase continuously.

Question. Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working. Answer the following :
Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer ?
Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason. 
Answer : . (a) Principle. When a current carrying coil is placed in magnetic field, it experiences a torque. Construction. It consists of a narrow rectangular coil PQRS consisting of a large number of turns of fine insulated copper wire wound over a frame made of light, non-magnetic metal. A soft iron cylinder known as the core is placed symmetrically within the coil and detached from it. The coil is suspended between the two cylindrical pole pieces (N and S of a strong permanent horse shoe magnet) by a thin flat phosphor bronze strip the upper end of which is connected to a movable torsion head T. The lower end of the coil is connected to a hair spring s of phosphor bronze having only a few turns. Redial magnetic field. The magnetic field in the small air gap between the cylindrical pole pieces is radial. The magnetic lines of force within the air gap are along the radii. On account of this, the plane of the coil remains always parallel to the direction of the magnetic field  (Image 9)
The magnetic field in the small air gap between the cylindrical pole-pieces is radial. On account of this, the plane of the coil remains always parallel to the direction of the magnetic field.
Theory : Let I = current flowing through the coil
B = magnetic field induction
l = length of the coil ;
b = breadth of the coil
N = number of turns in the coil
A (= l × b) = area of the coil
Moment of deflecting couple = NBIl × b = NBIA
When the coil deflects, the suspension fibre gets twisted. On account of elasticity, a restoring couple is set up in the fibre. This couple is proportional to the twist. If α be the angular twist then
Moment of restoring couple = kα
For equilibrium of the coil, NBIA = kα or α = NBAI/ k or α =KI
where K= NBA/ k is the galvanometer constant. Now, α ∝I or I ∝α (b) (i) By using soft iorn core . magnetic field is increased so sensitivity increases and mag. field becomes radial So angle between plane of coil & magnetic line of force is zero in all orientations of coil.
(b) (i) By using soft iorn core . magnetic field is increased so sensitivity increases and mag. field becomes radial So angle between plane of coil & magnetic line of force is zero in all orientations of coil.
(ii) Voltage sensitivity = Current sensitivity/ R
It means that V S increases if C S is increased. But if resistance of coil is also increases in same ratio then V S may be constant

Question. (a)Using Biot-Savart's law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop 
(b) What does a toroid consist of ? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside exterior to the toroid is zero.
Answer : . a. According to Biot-Savart's law, magnetic field due to a small element XY at point P is (Image10)
Resolving dB into two components :
(i) dB cos θ, which is perpendicular to the axis of the coil
(ii) dB sin θ which is along the axis of the coil and away from the centre of the coil. (Image10)
For any point inside the empty space surrounded by toroid and outside the toroid, magnetic field B is zero because the net current enclosed in these spaces is zero. But magnetic field is not exactly zero.

Question. (a)Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles.
(b)An α-particle and a proton are released from the centre of the cyclotron and made to accelerate.
(i)Can both be accelerated at the same cyclotron frequency ? Give reason to justify your answer.
(ii)When they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees ?
Answer : (Image 11)

1. Suppose a helical spring is suspended from the roof of a room and very small weight is attached to its lower end what will happen to the spring when a current is passed through it?Give reason to support your answer?

Ans Spring will contract due to the magnetic field produced by the turns of the coil and the weights will be lifted up.

2. One alpha particle and a deuteron entered perpendicularly in a uniform magnetic field with same velocity. Which one follow the greater circle?

Ans: As we know for a charge particle moving in a magnetic field, the radius of circular path: r = mv/qB 

As both the particles have same velocity therefore r_α/r_d = m_α q_d/m_d q_α 

4 X 1 / 2 X 2 = 1/1   Thus both particles will follow the same 

3. Out of Voltmeter and Millivoltmeter, which has the higher resistance?

Ans: We know the resistance connected to galvanometer to convert it into voltmeter is R = (V / Ig) - G So if R is higher, range of V will also be higher, so a Voltmeter has the higher resistance.

4. Proton is moving along the axis of a solenoid carrying current of 2 A and 50 number of turns per unit length. What will be the force acting on the particle. 

Ans: As the magnetic field produced by solenoid is always along its axis, so direction of velocity of proton is along the direction of field, therefore F = qvB Sin 0 = 0

5. Out of Ammeter and Milliammeter, which has the higher resistance?

Ans: We know the resistance connected to galvanometer to convert it into ammeter is S =(Ig/( I – Ig))xG So for higher resistance, the range of I should be small, therefore milliammeter has the higher resistance.

6.What will be the direction of magnetic field at 

magnetic moment of the arrangement=√ M_1²  +M₂²+2M₁M₂COSθ 

= √M²/4+M²/4+2M/2xM/2 cos90

=√ 2M²/4 =M/√2

29. A circular current carrying coil has a radius R. What is the distance from the centre of the coil on its axis where the magnetic field is 1/8 th of its value at the centre?

Ans: B_axial =1/8 B_centre

µ₀ IR² / 2(R²+r²)^3/2=  1x µ₀ I/ 8x2R

 (R²+r²)^3/2 =8R²

R²+ r²= 4R²

Hence , r =√ R

 

30.A magnetic needle suspended freely in a uniform magnetic field experiences torque but no net force. A nail made up of iron kept near a bar magnet experience  a force  of attraction and torque .Give reason.

Ans- Due to the non uniform magnetic field of bar magnet nail experience torque and translatory force.

 

What is the work done by a magnetic field on moving a charge? Give reason. Ans-   W= FScosθ  = FScos 90=0

 

32.A particle with charge q moving with velocity v in the plane of the paper enters a uniform magnetic field B acting perpendicular to the plane of the paper. Deduce an expression for the time period of the charge as it moves in a circular path in the field .

Why does the kinetic energy of the charge not change while moving in the magnetic field. Ans- particle moves in circular path

Bqv = mv²/r

r = mv/Bq

Time period T = 2 Π   r/v  =2  m/Bq 

 

33.A solenoid of length 0.6m has a radius of 1cm and is made up of 600 turns.It carries a current of 5A.What is the magnetic field inside and at ends of  solenoid.?

Ans- (i)At the centre

N=1000, B = µ₀ ni = 4 Π    x 10^-7 x 1000 x 5  = 6.2 x 10^-3 T (ii) At the ends

B = ½ µ₀ ni  = 3.1 x 10 ^-3 T 

 

34. An element dl = dx i  is placed at the origin and carries a large  current  I = 10A.What is the magnetic field on the y axis at a distance of 0.5m,

 

dB= µ₀ Idl Sin θ    /  4 Π   r² 

=10^-7 x 10 x 10^-2 /25 x 10^-2 

=4 x 10 ^-8 T 

Direction of dB is in +Z direction

 

35. You are given a copper wire carrying current I of length L. Now the wire is turned into circular coil. Find the number of turns in the coil so that the torque at the centre of the coil is to maximum.

 

Ans:   Let the number of turns be = n

Radius = r

Length=l

Length of the wire = circumference of n turns of coil

L = n  x 2 Π   r

 

r = L/2 Π   r

Maximum torque = nIBA  = nIB Π   r²

 

= nIB Π   (L/2 Π   n) ²

 

α   1/n

For maximum torque n should be minimum i.e.  n = 1.

36.What is the magnetic field produced at the centre of curvature of an arc of wire of radius r carrying current I subtends an angle Π   /2radians at its centre.

Ans:  B1 = B x θ  /2 Π   = (µ₀ I Π   /2r) x (2 Π   x2) 

B1= µ₀ I/8r

 

If B is the magnetic field produced at the centre of a circular coil of one turn of length L carrying current I then what is the magnetic field at the centre of the same coil which is made into 10 turns?

Ans : B1 = n² B = 10² B = 100  B.

 

38.A copper wire is bent into a square of each side 6cm.If a current of 2A is passed through a wire what is the magnetic field at the centre of the square?

Ans: B1 = 4 x µ₀ I/4 Π   a/2 ( sin 45  + sin 45 )

= 4 x µ₀ 2/4 Π   x3 ( 1/ 1.414 + 1/ 1.414  )

= 2 µ₀ /3 Π    ( 1/ 1.414 + 1/ 1.414  ) T

 

39.Find the magnetic moment of a wire of length l carrying current I bent in the form of a circle.

Ans: M = IA  = I x Π   r²

But l = 2 Π   r , i.e r = l/2 Π

M= Il²/4 Π

 

40.When current is flowing through two parallel conductors in the same direction they attract while two beams of electrons moving in the same direction repel each other. Why?

Ans-  Two conductors carrying current in same direction produce magnetic field and hence they attract.

While two electron beams moving in the same directions repel due to its electric field

(electrostatic force)

 

41.Draw diagrams to show behavior of magnetic field lines near a bar of (i) Alluminium (ii)

copper and (iii) mercury cooled to a very low temperature 4.2 K

Ans-  (i)Alluminium --- Paramagnetic

 

42.The hysteresis loss for a sample of 6 kg is 150 J/M²/cycle. If the density of iron is 7500 kg/m³, calculate the energy loss per hour at 40cycle.

Ans:  Volume of sample = mass/density = 6/7500 m³

Energy loss/cycle = energy loss per volume/cycle x( volume)=150x6/7500

Energy loss/sec = 150x6x40/7500

Energy loss /hour = 150x6x40x60x60/7500 J 

=1.728 x 104 J

 

43.A current carrying solenoid of 100 turns  has an area of cross section 10-4 m2 .When suspended freely through its centre, it can turn in a horizontal plane  .what is the magnetic moment of the solenoid for a current of 5A.Also calculate the net force and torque on solenoid if a uniform horizontal field of 10x10-2  T is set up at an angle of 30 degree with axis of solenoid when it is carrying the same current.

Ans:  M = nIA  = 100 x 5 x  10^-4 =500 x  10^-4  J/T Net force = 0

Torque = MB sinθ   = 5x10^-2  x 0.1x sin30 = 25 x 10^-4 Nm

H = R cos ð = .4x ^½   = 0.2 G

44.Two concentric circular coils A and B of radii 10 cm and 6 cm respectively, lie in the same vertical plane containing the north to south direction. coil A  has 30 turns and carries a current of 10 A . Coil B has 40 turns and carries a current of 15 A .the sense of the current in A   is anticlockwise and clockwise in B for an observer looking at the coils facing west. Give the magnitude and direction of net magnetic field

Ans: B due to the coils at the centre.

Coil A – 

R1 = 0.1m , n1=30,I1=10A 

B1 = µ₀ n1I1/2r1 = 6 Π    x 10_-4  T  directed towards east

 Coil B-

R2 = 0.6 m , n2=40, I2= 15 A

B2 = µ₀ n2I2/2r2 =2 Π    x 10 – T directed towards west

Net field B = B2 -  B1 = (20 Π -  6 Π    ) 10^-4  T towards west 

 

45.The vertical component of earth’s magnetic field at a given place is        3   times its horizontal component. If the total intensity of  earth’s magnetic field at a place is 0.4 G , find the value of horizontal component of earths field and angle of dip.

Ans: Tan ð = V/H = √3 ð = 60

As V = √3H and

B² = V² + H²  = 3H² + H² = 4H²

(0.4)² = 4H²    therefore

H = 0.2 G

 

46.An electron traveling west to east enters a chamber having a uniform electrostatic field in north to south direction.Specify the direction in which the uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Ans-  Due to the electrostatic field electron will be deflected towards north. To

keep it neutralized the magnetic force should deflect it towards south .For this purpose the magnetic field is to be applied perpendicular to the plane of the paper inward i.e vertically downward.

 

47. A straight horizontal conducting rod of length 0.5 m and mass 50 g is suspended by two vertical wires at its ends.A current of 5A is set up in the rod sdthrough the wires.(i) What magbnetric field should be set up normal to the conductor in order that  the tension in the wires is zero?(ii)What will be the tension in the wire if the direction of current is reversed keeping the magbetic field same as before?(neglect the mass if wure abd taje g=10m/s2 ) Ans:  (i) Magnetic force = weight

 

IlB sinθ   = mg

 

IlB = mg (θ  =90)

 

B =  mg/Il = 500x10^-3/2.5  = 200 x 10^-3 T

 

(ii)When the direction of fidl is reversed an additional force which was equal to weight of rod will be acting on the wires.

 

Net tension in wires = mg + mg  = 2 mg =2 x 50 x 10^-3 x 10  = 1 N

 

 

 

48.A   circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of

 

0.d10T normal to the plane of the coil.If the current in the coil is 5a,What is the (i)Total torque on the coil (ii) total force on the coil (iii) average dsforce on each electron  in the coil due to the magnetic field.(coil is made of copper, A= 10^-5  m2  ,free electron density in copper is 10²⁹ /m³)

 

Ans:  N= 20,  r= .1 m,  B = .1T, I = 5A

 

(i)Total torque = nIBAsin θ (θ  =0)

 

= 0

(ii)Total force on the coil is = 0, because force being equal and opposite and cancel eachother

(iii) Average force on electron = f = evB 

F = IB/nA = 5x 0.1/10²⁹ x 10^-5

=5 x 10 ^-25 N

V = I/neA

 

49. A Rowland ring of mean radius 15 cm has 3500 turns of wore wound on a ferromagnetic core  of  relative  permeability  800.What  is  the    magnetic  field  B  in  the  core  for  a magnetizing current of 1.2 A?

Ans: N=3500, r= 15 x 10^-2m

n = N/2 Π   r = 3500/2x3.14 x .15  = 3715.5 per m

µ₀ = 4 Π    x 10^-7 x TmA-1,  µr = 800 ,  I = 1.2

B = µ₀ µ_r nI = 4.48  T 

 

50 A straight wire of mass 200g and the length 1.5m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field B. What is the magnitude of B in tesla?

Ans: For equilibrium of the wire in mid – air, weight of the wire = force exerted by magnetic field

Mg =IlB sin90⁰

B = mg/Il =  200x 10^-3 x9.8/2x1.5 = 0.65T

 

51. A rigid circular loop of radius r and mass m lies in the x-y plane of a flat table and has a current I flowing in it. At this particular place the earth’s magnetic field is B = Bxi +Bzk. What is the value of I, so that loop starts tilting?

Ans:  M  = IA  =I r2k

B = Bxi +Bzk

Τ = M x  B = ( Iπ r2k) x(Bxi + Bzk)

= Iπ R2BXkxi = I  r2BXj

Torque due to the weight of the loop = mgr I  r²BX = mgr       Hence I = mg/ПrBX

 

52. In an ammeter, 10% of main current is passing through the galvanometer. If the resistance of the galvanometer is G, then what is the shunt resistance in ohms?

Ans: Ig=10% of I=0.1I

S=IgxG/I-Ig=0.1IG/I-0.1I=G/9

 

53. The two rails of arailway track insulated from each other and the ground is connected to a milli voltmeter. What is the reading g of the millivolmeter when the train passes at aspeed

180km/hr along the track, given that the vertical component of earth”s magnetic field is

0.2x10-4T and rails are separated by 1m e = Blv = 0.2x10^-4x1x180x5/18

= 10^-3V = 1Mv

 

54 A charged particle moving in a magnetic field penetrates a layer of lead and there by looses half of its kinetic energy.How does the radius of curvature of its path changes? Radius      r= mV/qB

Ans: If is the kinetic energy of the particle,then its momentum, p = mv√2mE_k

Radius , r =√2mE /Qb rα√E_k

This shows that K.E is halved, the radius is reduced to  1/√2 times its initial value.

 

55 The velocities of two α particles X and Y entering in an  uniform magnetic field are in the ratio 2:1.On entering the field ,they move in different circular paths .Give the ratio of the radii of their paths?

Ans:  qvB = mv²/r, r = mv/qB, rα v r_x/r_y = v_x/v_y = 2/1 = 2 

 

56 In an exercise to increase current sensitivity of a galvanometer by 25 % , its resisitance is increased by 1.5 times . How does the voltage sensititvity of the galvanometer be affected. Ans: I s ‘ = I s  + 25/100= 125/100 = 5/4 I s ---------  1

R’ = 1.5 R ----------- 2

V s + I s / R & Vs “ = Is’/R’ =5/4 I s/1.5 R

+ 5/6 V s

% decrease in voltage sensitivity

= (1-Vs’/V s) X 100

(1-5/6) X 100 = 16.7%