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Magnetic Properties of Matter

JEE Physics · 55 questions · Page 4 of 6 · Click an option or "Show Solution" to reveal answer

Q31
Statement I : The ferromagnetic property depends on temperature. At high temperature, ferromagnet becomes paramagnet. Statement II : At high temperature, the domain wall area of a ferromagnetic substance increases. In the light of the above statements, choose the most appropriate answer from the options given below :
A Statement I is false but Statement II is true
B Statement I is true but Statement II is false
C Both Statement I and Statement II are false
D Both Statement I and Statement II are true
Correct Answer
Option B
Solution

With increase in temperature domain volume decreases. Statement I true Statement 2 false

Q32
At a certain place the angle of dip is 30∘^\circ and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is :
A 13{1 \over {\sqrt 3 }}
B 12{1 \over 2}
C 3\sqrt 3
D 1
Correct Answer
Option A
Solution
BH=Bcos⁡30∘{B_H} = B\cos 30^\circ
⇒B=13G\Rightarrow B = {1 \over {\sqrt 3 }}G
Q33
The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 ×\times 10−-5. What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid?
A 1.2 ×\times 10−-5
B 1.2 ×\times 10−-3
C 1.8 ×\times 10−-3
D 2.4 ×\times 10−-5
Correct Answer
Option A
Solution
B′→=μ0(1+X)ni\overrightarrow {B'} = {\mu _0}(1 + X)ni

(in the material)

B′→=μ0ni\overrightarrow {B'} = {\mu _0}ni

(without material) So fractional increase is

B′−BB=X=1.2×10−5{{B' - B} \over B} = X = 1.2 \times {10^{ - 5}}
Q34
The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m, is : [Permeability of free space μ\mu0 = 4π\pi ×\times 10−-7 Wb/A-m]
A 4π\pi ×\times 10−-7
B 4π\pi ×\times 10−-4
C 4π\pi ×\times 10−-5
D 4π\pi ×\times 10−-6
Correct Answer
Option C
Solution

μ\mur = x + 1 = 99 + 1 = 100 ⇒\Rightarrow μ\mu = μ\murμ\mu0 = 100 ×\times 4μ\mu ×\times 10−-7 Wb/Am = 4μ\mu ×\times 10−-5 Wb/Am

Q35
A bar magnet having a magnetic moment of 2.0 ×\times 105 JT−-1, is placed along the direction of uniform magnetic field of magnitude B = 14 ×\times 10−-5 T. The work done in rotating the magnet slowly through 60∘^\circ from the direction of field is :
A 14 J
B 8.4 J
C 4 J
D 1.4 J
Correct Answer
Option A
Solution
U=−M→ . B→U = - \overrightarrow M \,.\,\overrightarrow B

So

Uf−Ui=−MB(1−cos⁡θ){U_f} - {U_i} = - MB(1 - \cos \theta )
=−14J= - 14J

So

W=−ΔU=14JW = - \Delta U = 14J
Q36
Given below are two statements : Statement I : Susceptibilities of paramagnetic and ferromagnetic substances increase with decrease in temperature. Statement II : Diamagnetism is a result of orbital motions of electrons developing magnetic moments opposite to the applied magnetic field. Choose the correct answer from the options given below :
A Both Statement I and Statement II are true.
B Both Statement I and Statement II are false.
C Statement I is true but Statement II is false.
D Statement I is false but Statement II is true.
Correct Answer
Option A
Solution

Statement I is true as susceptibility of ferromagnetic and paramagnetic materials is inversely related to temperature.

Statement II is true as because of orbital motion of electrons the diamagnetic material is able to oppose external magnetic field.

Q37
The soft-iron is a suitable material for making an electromagnet. This is because soft-iron has
A low coercivity and high retentivity.
B low coercivity and low permeability.
C high permeability and low retentivity.
D high permeability and high retentivity.
Correct Answer
Option C
Solution

Electromagnet requires high permeability and low retentivity.

Q38
An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 ×\times 10−-4 Wbm−-2. The frequency of revolution of the electron will be : (Take mass of electron = 9.0 ×\times 10−-31 kg)
A 1.6×105 Hz1.6 \times 10^{5} \mathrm{~Hz}
B 5.6×105 Hz5.6 \times 10^{5} \mathrm{~Hz}
C 2.8×106 Hz2.8 \times 10^{6} \mathrm{~Hz}
D 1.8×106 Hz1.8 \times 10^{6} \mathrm{~Hz}
Correct Answer
Option C
Solution
T=2πmBqT = {{2\pi m} \over {Bq}}

⇒\Rightarrow Frequency

f=Bq2πmf = {{Bq} \over {2\pi m}}
=10−4×1.6×10−192π×9×10−31= {{{{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}}} \over {2\pi \times 9 \times {{10}^{ - 31}}}}
≃2.8×106\simeq 2.8 \times {10^6}

Hz

Q39
A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P\mathrm{P} of dip⁡30∘\operatorname{dip} 30^{\circ}. The number of oscillations per minute become 10 at another place Q\mathrm{Q} of 60∘60^{\circ} dip. The ratio of the total magnetic field at the two places (BQ:BP)\left(B_{Q}: B_{P}\right) is :
A 3:4\sqrt{3}: 4
B 4:34: \sqrt{3}
C 3:2\sqrt{3}: 2
D 2:32: \sqrt{3}
Correct Answer
Option A
Solution
T∝1Bcos⁡δT \propto {1 \over {\sqrt {B\cos \delta } }}
⇒T1T2=B2cos⁡δ2B1cos⁡δ1\Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{B_2}\cos {\delta _2}} \over {{B_1}\cos {\delta _1}}}}
⇒3 s6 s=B1B2×1232\Rightarrow {{3\,s} \over {6\,s}} = \sqrt {{{{B_1}} \over {{B_2}}} \times {{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}}
⇒B2B1=(12)2×3=34\Rightarrow {{{B_2}} \over {{B_1}}} = {\left( {{1 \over 2}} \right)^2} \times \sqrt 3 = {{\sqrt 3 } \over 4}
Q40
The vertical component of the earth's magnetic field is 6×10−5 T6 \times 10^{-5} \mathrm{~T} at any place where the angle of dip is 37∘37^{\circ}. The earth's resultant magnetic field at that place will be (\left(\right.Given tan⁡37∘=34)\left.\tan 37^{\circ}=\dfrac{3}{4}\right)
A 8×10−5 T8 \times 10^{-5} \mathrm{~T}
B 6×10−5 T6 \times 10^{-5} \mathrm{~T}
C 5×10−4 T5 \times 10^{-4} \mathrm{~T}
D 1×10−4 T1 \times 10^{-4} \mathrm{~T}
Correct Answer
Option D
Solution

∴\therefore

sin⁡37∘=BvBnet\sin 37^\circ = {{{B_v}} \over {{B_{net}}}}
⇒Bv=Bnetsin⁡37∘\Rightarrow {B_v} = {B_{net}}\sin 37^\circ
⇒Bnet=Bvsin⁡37∘\Rightarrow {B_{net}} = {{{B_v}} \over {\sin 37^\circ }}
=6×10−535= {{6 \times {{10}^{ - 5}}} \over {{3 \over 5}}}
=10×10−5 T= 10 \times {10^{ - 5}}\,T
=1×10−4 T= 1 \times {10^{ - 4}}\,T
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