Class 12-science H C VERMA Solutions Physics Chapter 15 - Magnetic Properties of Matter
Magnetic Properties of Matter Exercise 286
Solution 1
Here given in
question,
Current in the solenoid, I =
2 A
Magnetic intensity at the centre of long solenoid, H = 1500 Am-1
Magnetic field produced by a solenoid(B) is given byB = μ0ni ...(1)
where, n = number of
turns per unit length
i =
electric current through the solenoid
Also, the relation between magnetic field strength(B) and magnetic intensity(H) is given
byH = 
...(2)
From equations (1) and (2), we get:H = ni⇒ 1500
A/m = n × 2⇒ n = 7.5 turns/cm
The number of turns per centimetre of the solenoid is 7.5.
Solution 2
Given: (a)
Intensity of magnetisation H = 1500 A/m As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, the end effects may be neglected. The sole effect of the rod in the magnetic field of the solenoid is that a magnetisation will be induced in the rod depending on rod magnetic properties. There is no effect of the rod on the magnetic intensity at the centre.
(b) Magnetisation of the core is I = 0.12 A/m As we know that
I=XH …where χ is the
susceptibility of the material of the rod.∴ X=
=0.12/1500
∴ X =8 × 10-5
(c) The material is paramagnetic.
Solution 3
Given that, Magnetic field strength without iron core is B1 = 2.5×10-3 T Magnetic field after introducing the iron core is B2 = 2.5 T Area of cross-section of the iron core is A = 4×10-4 m2 Number of turns per unit length n = 50 turns/cm = 5000 turns/m (a) Magnetic field produced by a solenoid (B) is given by, B =μ0ni where i = electric current in the solenoid 2.5 × 10-3 = 4ππ × 10-7 × 5000 × i⇒ i=2.5×10-3/(4π×10-7×5000)
=0.398 A=0.4 A
(b) Magnetisation (I) is given by
I=B/u0 -H, ……where B is the net magnetic field after introducing the core,
i.e. B=2.5 T
And μ0H will be the magnetising field, i.e. the diffrence between the two magnetic fields' strengths.
⇒ I=2.5 × 10-3/4π × 10-7 *(B2-B1)
⇒ I ≈ 2 × 106 A/m
(c) Intensity of magnetisation(I) is given by,
I = M/V
⇒ I= m × 2I/A × 2I = m/A
⇒ m =lA ⇒ m= 2 × 106 × 4 × 10-4⇒ m= 800 A-m
Solution 4
(a) Given:
Distance of the observation point from the centre of the bar magnet, d=15cm=0.15m
Length of the bar magnet, l=1cm=0.01m
Are of cross-section of the bar magnet, A=1.0 cm2=1×10-4 m2
Magmatic filed strength of the bar magnet, B = 1.5 × 10-4 T
As the observation point lies at the end-on position, magnetic field (B) is given by,
On substituting the respective values, we get:
=2.5A
(b) Intensity of magnetization (I) is given by,
= 2.5 × 106 A/m
(c)
Net H = HN+Hs
= 884.6=8.846 × 102
= 314 T
= 𝜋 × 10-7(2.5×106 + 2 × 884.6)
=3.14 T
Solution 5
Susceptibility of annealed iron is χ = 5500 The relation between permeability and susceptibility is given by Permeability, μ = μ0(1 + x)μ = 4ππ × 10-7 (1 + 5500)⇒ μ = 4 × 3.14 × 10-7 × 5501⇒ μ = 69092.56 × 10-7⇒ μ = 6.9 × 10-3
Solution 6
Here given that, Magnetic field strength is B = 1.6 TMagnetising intensity in a material is H = 1000 A/m The relation between magnetic field and magnetising field is given byμ=B/H
⇒ μ=1.6/1000
⇒ μ=1.6×10-3
Relative permeability is defined as the ratio of permeability in a medium to that in vacuum.
So, μr= μ/ μ0=1.6 × 10-3/ 4π×10-7
⇒ μr=0.127 × 104
⇒ μr=1.3 × 104
Relative permeability (μr) is given by,
μr=(1+χ)
⇒ X= 1.3 × 103 -1
⇒ X= 1300-1
⇒ X= 1299
= 1.299 × 103
⇒ X ≈1.3 × 103
The susceptibility χ of the material is 1.3 × 103
Magnetic Properties of Matter Exercise 287
Solution 7
Given, Susceptibility of magnesium at 300 K is χ1=1.2×10-5 Let T1 be the temperature at which susceptibility of magnesium is 1.2×10-5 and T2 be the temperature at which susceptibility of magnesium is 1.8×10-5. According to Curie's law, χ = C/T, where C is Curie's constant.⇒ X1/X2=T2/T2
⇒ 1.2 × 105/1.8 × 10-5=T2/300
⇒ T2=12/18 × 300
⇒ T2=200 K
At 200K temperature will the susceptibility increase.
Solution 8
Given that in problem, No of atoms per unit volume is f = 8.52 × 1028 atoms/m3Magnetisation per atom is M = 2 × 9.27 × 10-24 A-m2 (a) Intensity of magnetisation, I = M,V⇒ I=2 × 9.27 × 10-24 × 8.52 × 1028⇒ I=1.58 × 106 A/m.
(b) For maximum magnetisation, the magnetising field will be equal to the intensity of magnetisation. So, I = H Magnetic field (B) will be, B = 4ππ × 10-7 × 1.58 × 106 ⇒ B ≈ 19.8 × 10-1 = 2.0 T.
Solution 9
Given that, Number of turns per unit length is n = 40 turns/cm = 4000 turns/m Magnetising field is H = 4×104 A/m Magnetic field inside a solenoid (B) is given by,B = μ0nI, where, n = number of turns per unit length. I = current through the solenoid. ∴ B/μ0= nI = H ⇒ H=NI/l
⇒ I=Hl/N = H/n
⇒ I=4 × 104/4000
=10 A
So, the current is 10A.
