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Magnetic Properties of Matter

JEE Physics · 55 questions · Page 1 of 6 · Click an option or "Show Solution" to reveal answer

Q1
The coercivity of a small magnet where the ferromagnet gets demagnetized is 3×103 Am−1.3 \times {10^3}\,A{m^{ - 1}}. The current required to be passed in a solenoid of length 1010 cmcm and number of turns 100,100, so that the magnet gets demagnetized when inside the solenoid, is :
A 3030 mAmA
B 6060 mAmA
C 33 AA
D 6A6A
Correct Answer
Option C
Solution

Magnetic field in solenoid

B=μ0niB = {\mu _0}ni
⇒Bμ0=ni\Rightarrow {B \over {{\mu _0}}} = ni

(Where

n=n=

number of turns per unit length)

⇒Bμ0=NiL\Rightarrow {B \over {{\mu _0}}} = {{Ni} \over L}
⇒3×103=100i10×10−2\Rightarrow 3 \times {10^3} = {{100i} \over {10 \times {{10}^{ - 2}}}}
⇒i=3A\Rightarrow i = 3A
Q2
A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 ms−1. The earth’s magnetic field over Delhi is 5 × \times 10−5 T with the declination angle ~ 0o and dip of θ\theta such that sin θ\theta = 2/3 . If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to :
A VB = 45 mV; VW = 120 mV with right side of pilot at higher voltage
B VB = 45 mV; VW = 120 mV with left side of pilot at higher voltage
C VB = 40 mV; VW = 135 mV with right side of pilot at high voltage
D VB = 40 mV; VW = 135 mV with left side of pilot at higher voltage
Correct Answer
Option B
Solution

VB = vhBcosθ\theta = 240 ×\times 5 ×\times 5 ×\times 10−-5×\times

53{{\sqrt 5 } \over 3}

= 44.7 ×\times 10−-3 V = 45 mV Vw =

ll

vB sinθ\theta = 15 ×\times 240 ×\times 5 ×\times 10−-5 ×\times

23{2 \over 3}

= 1200 ×\times 10−-4 V = 120 mV From right hand rule, the charge moves to the left side of the pilot.

Q3
A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is : (1G = 10−-4 T)
A 2.880 ×\times 102 J T−-1
B 2.880 J T−-1
C 2.880 ×\times 103 J T−-1
D 28.80 J T−-1
Correct Answer
Option B
Solution
B=2B0sin⁡θB = 2{B_0}\sin \theta
B=2μ04πmr2×7rB = 2{{{\mu _0}} \over {4\pi }}{m \over {{r^2}}} \times {7 \over r}
⇒0.4×10−4=2×10−7×m×7(72+182)3/2×104\Rightarrow 0.4 \times {10^{ - 4}} = 2 \times {10^{ - 7}} \times {{m \times 7} \over {{{({7^2} + {{18}^2})}^{3/2}}}} \times {10^4}

∴\therefore

m=4×10−2×(373)3/214m = {{4 \times {{10}^{ - 2}} \times {{(373)}^{3/2}}} \over {14}}

⇒\Rightarrow M = m ×\times 14 cm =

=m×14100= m \times {{14} \over {100}}

⇒\Rightarrow

M=0.04×(373)3/214×14100M = {{0.04 \times {{(373)}^{3/2}}} \over {14}} \times {{14} \over {100}}

⇒\Rightarrow M = 2.880 J/T

Q4
A magnetic needle of magnetic moment 6.7 ×\times 10-2 A m2 and moment of inertia 7.5 ×\times 10-6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:
A 8.76 s
B 6.65 s
C 8.89 s
D 6.98 s
Correct Answer
Option B
Solution

Given : Magnetic moment, M = 6.7 × 10–2 Am2 Magnetic field, B = 0.01 T Moment of inertia, I = 7.5 × 10–6 Kgm2 Using, T =

2πIMB2\pi \sqrt {{I \over {MB}}}

=

2π7.5×10−66.7×10−2×0.012\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}}

=

2π10×1.06{{2\pi } \over {10}} \times 1.06

s Time taken for 10 complete oscillations t = 10T = 2π\pi × 1.06 = 6.6568

≃\simeq

6.65 s

Q5
A paramagnetic material has 1028 atoms/m3. Its magnetic susceptibility at temperature 350 K is 2.8 × \times 10–4. Its susceptibility at 300 K is :
A 3.726 × \times 10–4
B 2.672 × \times 10–4
C 3.672 × \times 10–4
D 3.267 × \times 10−-4
Correct Answer
Option D
Solution

x α\alpha

1TC{1 \over {{T_C}}}

curie law for paramagnetic substane

x1x2{{{x_1}} \over {{x_2}}}

=

TC2TC1{{{T_{{C_2}}}} \over {{T_{{C_1}}}}}
2.8×10−4x2=300350{{2.8 \times {{10}^{ - 4}}} \over {{x_2}}} = {{300} \over {350}}

x2 =

2.8×350×10−4300{{2.8 \times 350 \times {{10}^{ - 4}}} \over {300}}

= 3.266 ×\times 10−-4

Q6
A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45o, and 40 times per minute where the dip is 30o. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1B2{{{B_1}} \over {{B_2}}} is best given by :
A 1.8
B 2.2
C 0.7
D 3.6
Correct Answer
Option C
Solution
f1=12πμB1cos⁡45oI{f_1} = {1 \over {2\pi }}\sqrt {{{\mu {B_1}\cos {{45}^o}} \over I}}
f2=12πμB2cos⁡30oI{f_2} = {1 \over {2\pi }}\sqrt {{{\mu {B_2}\cos {{30}^o}} \over I}}
f1f2=B1cos⁡45oB1cos⁡30o{{{f_1}} \over {{f_2}}} = {{{B_1}\cos {{45}^o}} \over {{B_1}\cos {{30}^o}}}

∴\therefore

B1B2=0.7{{{B_1}} \over {{B_2}}} = 0.7
Q7
A magnetic needle lying parallel to a magnetic field requires WW units of work to turn it through 60∘.{60^ \circ }. The torque needed to maintain the needle in this position will be :
A 3 W\sqrt 3 \,W
B WW
C 32W{{\sqrt 3 } \over 2}W
D 2W2W
Correct Answer
Option A
Solution
W=MB(cos⁡θ1−cos⁡θ2)W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)
=MB(cos⁡θ∘−cos⁡60∘)= MB\left( {\cos {\theta ^ \circ } - \cos {{60}^ \circ }} \right)
=MB(1−12)=MB2= MB\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}

∴\therefore

τ=MB sin⁡θ=MB sin⁡ 60∘\tau = MB\,\sin \theta = MB\,\sin \,{60^ \circ }
=3MB2=3W= \sqrt 3 {{MB} \over 2} = \sqrt 3 W
Q8
Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3 s3 \mathrm{~s} and 4 s4 \mathrm{~s} respectively. If their moments of inertia are in the ratio of 3:23: 2, then the ratio of their magnetic moments will be:
A 2 : 1
B 8 : 3
C 1 : 3
D 27 : 16
Correct Answer
Option B
Solution
T=2πIMBHT = 2\pi \sqrt {{I \over {M{B_H}}}}
⇒T1T2=I1I2M2M1\Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}} \sqrt {{{{M_2}} \over {{M_1}}}}
⇒34=32M2M1\Rightarrow {3 \over 4} = \sqrt {{3 \over 2}} \sqrt {{{{M_2}} \over {{M_1}}}}
⇒M1M2=32×169=83\Rightarrow {{{M_1}} \over {{M_2}}} = {3 \over 2} \times {{16} \over 9} = {8 \over 3}
Q9
A thin rectangular magnet suspended freely has a period of oscillation equal to T.T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T′,T', the ratio T′T{{T'} \over T} is
A 122{1 \over {2\sqrt 2 }}
B 12{1 \over 2}
C 22
D 14{1 \over 4}
Correct Answer
Option B
Solution

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

T=2πIμBHT = 2\pi \sqrt {{I \over {\mu {B_H}}}}

where

I=I=

Moment of inertia of the rectangular magnet

μ=\mu =

Magnetic moment

BH={B_H} =

Horizontal component of the earth's magnetic field Case 1 :

T=2πIμBHT = 2\pi \sqrt {{I \over {\mu {B_H}}}}

where

I=112Mℓ2I = {1 \over {12}}M{\ell ^2}

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

T′=2πI′μ′BHT' = 2\pi \sqrt {{{I'} \over {\mu '{B_H}}}}

where

I′=112(M2)(ℓ2)2=I8I' = {1 \over {12}}\left( {{M \over 2}} \right){\left( {{\ell \over 2}} \right)^2} = {I \over 8}

and

μ′=μ2\mu ' = {\mu \over 2}

∴\therefore

T′T=I′μ′×μI{{T'} \over T} = \sqrt {{{I'} \over {\mu '}} \times {\mu \over I}}
=I/8μ/2×μI= \sqrt {{{I/8} \over {\mu /2}} \times {\mu \over I}}
=14=12= \sqrt {{1 \over 4}} = {1 \over 2}
Q10
The magnetic lines of force inside a bar magnet
A are from north-pole to south-pole of the magnet
B do not exist
C depend upon the area of cross-section of the bar magnet
D are from south-pole to north-pole of the Magnet
Correct Answer
Option D
Solution

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.

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