CBSE Class 12 Physics HOTs Magnetic Effects of Current

ONE MARK QUESTIONS

Question. Relative permeability of a material µr = 0.5. Identify the nature of the magnetic material and write its relation of magnetic susceptibility.
Answer : The nature of Magnetic Material is Diamagnetic.
µr=1+χm

Question. What are permanent magnets? Give one example?
Answer : Permanent Magnet are those magnets which have high retentivity and coercivity. The magnetision of permanent magnet is not easily destroyed even if it is handled roughly or exposed in stray, reverse magnetic field ,e.g: steel.

Question. Where on the surface of earth’s vertical component of earth’s magnetic field zero?
Answer : At equator, vertical component will be zero

Question. The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°.what is the value of vertical component of the earth’s magnetic field at equator?
Answer : {The vertical and horizontal components of the earth’s magnetic fields are perpendicular to each other}
Horizontal component of earth’s magnetic field,
H = Be cos 60°= B{given}
= Be * ½ = B
Or Be = 2B
Vertical component of earth’s magnetic field,
V = Be sin 60°
V =2B *√3/2
V = √3B

Question. What is the angle of dip at a place where the horizontal component and vertical components of the earth’s magnetic field are equal?
Answer : The angle of dip is given by
Δ = tan⁻1 {Bᵥ / BH}
Bᵥ = vertical component of the earth’s magnetic field.
BH= horizontal component of the earth’s magnetic field.
So, as Bᵥ= BH
Then, δ = tan-1[1] = 45°
Therefore, The angle of dip will be δ = 45°.

Question. A magnetic needle free to rotate in vertical plane and tends itself vertically at a certain place on the earth what are the values of
a. Horizontal component of the earth’s magnetic field and
b. angle of dip at this place ?
Answer : a. The coil is free to move in vertical plane , it means that there is no component A the earth’s magnetic field in horizontal direction, so the horizontal component of the earth’s magnetic field is 0
b. The angle of dip is 90°

Question. Where on the surface of the earth is the angle of dip 90°?
Answer : At the poles, the angle o dip is 90°

Question. The permeability of a magnetic material is 0.9983. Name the type of magnetic material, it represents.
Answer : The magnetic material is Diamagnetic substance for which µr<1.

Question. The susceptibility of a magnetic material is 1.9 * 10⁻⁵. Name the magnetic material, it represents.
Answer : The small and positive susceptibility of 1.9 *10⁻⁵ represents paramagnetic substance.

Question. The susceptibility of a magnetic material is -4.2 * 10⁻⁶.Name the type of magnetic material, it represents.
Answer : Negative susceptibility diamagnetic substance

Question. What is the characteristics property of a diamagnetic material?
Answer : Diamagnetic material acquires magnetization in the opposite direction of the magnetic field when they are placed in an external magnetic field.

Question. Define the term magnetic declination.
Answer : Magnetic declination the angle between geographical meridian and magnetic meridian at any place of the earth is known as magnetic declination.

TWO MARKS QUESTIONS

Question. An electron moves around the nucleus in a hydrogen atom of radius 0.5 Aͤ, with a velocity of 2 * 10⁶m/s. Calculate the following
(1) the equivalent current due to orbital motion of electron
(2) the magnetic field produced at the centre of the nucleus
(3) the magnetic moment associated with the electron.
Answer : Here r = 0.51 * 10⁻10 m, v = 2 * 105 ms-1
(1) I = e/T = ev/2πr = 1.6 * 10-19 * 2 *105 /2π * 0.51 * 10-10 = 10-4A
(2) B = μ0 I/2r = 4π * 10-7 * 10-4 /2 *0.51 *10-10 =1.23 T
(3) M = IA =ev/2πr * πr = evr /2
1.6 * 10-19*2*105*0.51*10-10 =8.16 * 10-25 Am2

Question. Out of the two magnetic materials, A has relative permeability slightly greater than unity while B has less than unity. Identify the nature of the material A and B. Will their susceptibilities be positive or negative?
Answer : The nature of material A is paramagnetic and its susceptibility ᵡM is positive
The nature of the material B is Diamagnetic and its susceptibility ᵡM is negative.

Question. Give two points to distinguish between a paramagnetic and diamagnetic substance.
Answer : 

Question. Write two properties of a material which makes it suitable for making electromagnet?
Answer : An electromagnet consists of a core, made of a ferromagnetic material placed inside a solenoid. it behaves like a strong magnet when current flows through the solenoid and effectively loses its magnetism when the current is switched off.
a. A permanent magnet is also made up of a ferromagnetic material but it retains its magnetism at room temperature for a long time after being magnetized one.
b. Properties of material are as below:
(i) High permeability
(ii) Low permeability
(iii) Low coercivity

Question. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties.
Answer : Ferromagnetic substances as these substances have very high magnetic permeability.
Properties (1) High retentivity (2) High susceptibility

Question. a. How does a diamagnetic material behave when it is cooled to very low temperature?
b. Why does a paramagnetic sample display greater magnetisation when cooled? Explain.
Answer : a. As the resistance (electrical of metal decreases with decrease in temperature) nut for diamagnetic substances, the variation of susceptibility is very small, i.e. diamagnetic material are unaffected by the change in temperature (except bismuth).
b. Paramagnetic materials when cooled due to thermal agitation tendency alignment of magnetic dipoles decreases. Hence, they show greater magnetisation.

Question.A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its North tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer : Angle of dip, δ = 60° = π/3
Horizontal component of the earth’s magnetic field, H =0.4 G
Earth magnetic field (Bₑ) =?
Horizontal component of the earth’s magnetic field, H = Bₑ cos δ
Bₑ = H/cos δ = 0.4 G /cos 60° = 0.4 G /½ = 0.8 G
Therefore, Bₑ = 0.8 G

Question. Distinguish between diamagnetic and ferromagnetic materials in terms of
(I) Susceptibility and
(II) Their behavior in a non uniform magnetic field.
Answer : a. Susceptibility for diamagnetic material it is independent of magnetic field and temperature (except for bismuth at low temperature)
Susceptibility for ferromagnetic material The susceptibility of ferromagnetic materials decreases steadily with increase in temperature. At the Curie temperature, the ferromagnetic materials become paramagnetic.
b. Behavior in non uniform magnetic field Diamagnets are feebly repelled ,whereas ferromagnets are strongly attracted by non uniform field, i.e. diamagnets move in the direction of decreasing field, whereas ferromagnet feels force in the direction of increasing field intensity.

Question(a) Write two characteristics of a material used for making permanent magnets?
(b) Why is core of electromagnets made of ferromagnetic materials?
Answer : (a) Two characteristics of material used for making permanent magnets are
(i) high coercivity (ii) high retentivity
(b) Core of an electromagnet made of ferromagnetic material because of its
(i) low coercvity (ii) low hysterisis loss (ii) low retentivity

Question. The horizontal component of the earth’s magnetic fields at a place is 3 times its vertical component there. Finds the value of the angle of dip at that place? What is the ratio of the horizontal component to the total magnetic field of the earth at that place?
Answer : As vertical and horizontal components of magnetic fields are perpendicular to each other,when their magnitudes are equal,resultant will divide their angle equally.
According to the question,
H =√3V
Where, H and V are the horizontal and vertical components of the earth’s magnetic field. if angle of dip at the place is δ, then
Tan δ = V/H = V/√3V
Tan δ = 1/√3
δ = π/6
Horizontal component of the earth’s magnetic field,
H =Bₑ cos δ
Bₑ =Earth’s magnetic field
H/Bₑ = cos δ
= cos π/6
=√3/2
H : Bₑ = √3 :2