1.Spring will contract due to the magnetic field produced by the turns of the coil and the weights will be lifted up.
2. 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 rα/rd = mα qd/md qα
Thus both particles will follow the same
3. 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. 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. 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. The magnetic field due to AB and EF is as the direction of length vector is along the radius vector,
7. As MCG detect only the average value of current and the average value of AC for a complete cycle is zero. Therefore MCG can not detect AC in a circuit.
8. For a given length, the circle has the greatest area, as
τ = NIAB
i.e. torque is proportional to area, so circular current loop experiences the greater torque.
9. The force will be zero as the stationary charge particle does not produce any force.
10. r = mv/qB ----------- (i)
By equ (i) and equ (ii)
As the radius is
proportional to square root of kinetic
energy, so if the kinetic energy is halved the radius become √1/2 times of its
11.As the resistance of V is very high so the effective resistance of circuit become very high, so the current flows in circuit is extremely low therefore the deflection is almost zero, while the V measures the potential difference between the points so it shows the reading due to battery.
13. Temperature in the core of earth is higher than Curie temperature of Iron.
16. Magnitude of Earth’s magnetic field is much smaller than magnitude of the field produced by poles of galvanometer.
18. θ = 0 (Dipole is parallel to field.)
19. Becomes Paramagnetic.
20. M1 = M/2, M=M/2
22. Nil, because the number of magnetic lines entering the surface is equal to the number of lines going out of it.
23. No Ans
25. The fields of the two wires will be in the opposite directions at the midway point.
B =B1 –B2 =μ0I/2πr -μ0I/2 π r =0
26. Kinetic Energy gained=qv=ex1Kv=1keV
27. φ =B.A = μo I Πr 2 / 2r
= μoIRΠ / 2
ie φ α R
28. magnetic moment of the arrangement=√ M1
= √M2/4+M2/4+2M/2xM/2 cos90
=√ 2M2/4 =M/√2
29. Baxial =1/8 Bcentre
μo IR 2 /
2(R2+r2)3/2 = 1x μo I / 8x2R
R2+r2 = 4R2
Hence , r =√ R
30. Due to the non uniform magnetic field of bar magnet nail experience torque and translatory force.
31. W= FScosq = FScos 90=0
32. particle moves in circular path
Bqv = mv2/r
r = mv/Bq
Time period T = 2 π r/v =2 m/Bq
33. (i)At the centre
N=1000, B = μo ni = 4 π x 10-7 x 1000 x 5 = 6.2 x 10-3 T
(ii) At the ends
B = ½ μo ni = 3.1 x 10 -3 T
34. dB= μo Idl Sin q / 4 P 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. Let the number of turns be = n
Radius = r
Length of the wire = circumference of n turns of coil
L = n x 2π r
r = L/2 π r
Maximum torque = nIBA = nIB π r2
= nIB π (L/2 π n) 2
For maximum torque n should be minimum
i.e. n = 1.
36. B1 = B x q /2π = (μo I π /2r) x (2 π x2)
B1= μo I/8r
37. B1 = n2 B = 102 B = 100 B.
38. B1 = 4 x μo I/4π a/2 ( sin 45 + sin 45 )
= 4 x μo 2/4 π x3 ( 1/ 1.414 + 1/ 1.414 )
= 2 μ0o /3 π ( 1/ 1.414 + 1/ 1.414 ) T
39. M = IA = I xπ r2
But l = 2 π r , i.e r = l/2 P
40. 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. (i)Alluminium --- Paramagnetic
(ii) Copper and mercury = diamagnetic
42. 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. M = nIA = 100 x 5 x 10-4 =500 x 10-4 J/T
Net force = 0
Torque = MB sin θ = 5x10-2 x 0.1x sin30 = 25 x 10-4 Nm
H = R cos ð = .4x ½ = 0.2 G
44. B due to the coils at the centre.
Coil A –
R1 = 0.1m , n1=30,I1=10A
B1 = μo n1I1/2r1 = 6 π x 10-4 T directed towards east
Coil BR2 -
= 0.6 m , n2=40, I2= 15 A
B2 = μo 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. 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. 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. (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. N= 20, r= .1 m, B = .1T, I = 5A
(i)Total torque = nIBAsin θ (θ =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
V = I/neA
F = IB/nA = 5x 0.1/1029 x 10-5
=5 x 10 -25 N
49. N=3500, r= 15 x 10-2m
n = N/2 π r = 3500/2x3.14 x .15 = 3715.5 per m
μo = 4 π x 10-7 x TmA-1, μr = 800 , I = 1.2
B = μo μr nI = 4.48 T
50. 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. 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. Ans: Ig=10% of I=0.1I
53. NO Ans
54. If is the kinetic energy of the particle,then its momentum, p = mv√2mEk
Radius , r =√2mE /Qb
This shows that K.E is halved, the radius is reduced to 1/√2 times its initial value.
55. qvB = mv2/r, r = mv/qB, rα v
rx/ry = vx/vy = 2/1 = 2
56. 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%
cbse physics (chapter wise with hint / solution) class xii (by mr. sreekumaran nair)