Practice Start Test

Electromagnetic Induction

JEE Physics · 59 questions · Page 1 of 6 · Click an option or "Show Solution" to reveal answer

Q1
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon
A the rates at which currents are changing in the two coils
B relative position and orientation of the two coils
C the currents in the two coils
D the materials of the wires of the coils
Correct Answer
Option B
Solution

Mutual conductance depends on the relative position and orientation of the two coils.

Q2
An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 ×\times 10−-4 Wb/m2 and the angle of dip is 60∘^\circ. The emf induced between the tips of the plane wings will be __________.
A 88.37 mV
B 62.50 mV
C 54.125 mV
D 108.25 mV
Correct Answer
Option D
Solution
ε\varepsilon

ind = (Bv) LV and Bv = BTotal sin60o ∴\therefore

ε\varepsilon

ind = (2.5 ×\times 10−-4)(sin 60o) ×\times 10 ×\times 180 ×\times

518{5 \over {18}}

= 108.25 mV

Q3
A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.
A 20 V
B 8 V
C 16 V
D 12 V
Correct Answer
Option C
Solution

Induced emf across the ends = Bvll = 2 × 8 × 1 = 16 V

Q4
A conducting circular loop of radius 10π\dfrac{10}{\sqrt\pi} cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :
A emf = 10 mV
B emf = 5 mV
C emf = 100 mV
D emf = 1 mV
Correct Answer
Option A
Solution

EMF=dϕdt=BA−0tA=πr2=π(0.12π)=0.01 B=0.5EMF=(0.5)(0.01)0.5=0.01 V=10 mV\begin{aligned} & \mathrm{EMF}=\dfrac{\mathrm{d} \phi}{\mathrm{dt}}=\dfrac{\mathrm{BA}-0}{\mathrm{t}} \\\\ & \mathrm{A}=\pi \mathrm{r}^2=\pi\left(\dfrac{0.1^2}{\pi}\right)=0.01 \\\\ & \mathrm{~B}=0.5 \\\\ & \mathrm{EMF}=\dfrac{(0.5)(0.01)}{0.5}=0.01 \mathrm{~V}=10~ \mathrm{mV}\end{aligned}

Q5
The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -
A 740 J
B 637.5 J
C 540 J
D 437.5 J
Correct Answer
Option D
Solution
Ldidt=25L{{di} \over {dt}} = 25
L×151=25L \times {{15} \over 1} = 25
L=53HL = {5 \over 3}H
ΔE=12×53×(252−102)\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})
=56×525=437.5= {5 \over 6} \times 525 = 437.5

J

Q6
A metal conductor of length 11 mm rotates vertically about one of its ends at angular velocity 55 radians per second. If the horizontal component of earth's magnetic field is 0.2×10−4T,0.2 \times {10^{ - 4}}T, then the e.m.f.e.m.f. developed between the two ends of the conductor is
A 5mV5mV
B 50μV50\mu V
C 5μV5\mu V
D 50mV50mV
Correct Answer
Option B
Solution
ℓ=1m,  ω=5 rad/s,  B=0.2×10−4T\ell = 1m,\,\,\omega = 5\,rad/s,\,\,B = 0.2 \times {10^{ - 4}}T
ε=Bωℓ22=0.2×10−4×5×12=50μV\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V
Q7
Which of the following units denotes the dimension ML2Q2{{M{L^2}} \over {{Q^2}}}, where QQ denotes the electric charge?
A Wb/m2Wb/{m^2}
B Henry (H)(H)
C H/m2H/{m^2}
D Weber (Wb)(Wb)
Correct Answer
Option B
Solution

Mutual inductance

=ϕI=BAI= {\phi \over I} = {{BA} \over I}

[Henry]

=[MT−1Q−1L2][QT−1]=ML2Q−2= {{\left[ {M{T^{ - 1}}{Q^{ - 1}}{L^2}} \right]} \over {\left[ {Q{T^{ - 1}}} \right]}} = M{L^2}{Q^{ - 2}}
Q8
A boat is moving due east in a region where the earth's magnetic fields is 5.0×10−55.0 \times {10^{ - 5}} NA−1 m−1N{A^{ - 1}}\,{m^{ - 1}} due north and horizontal. The best carries a vertical aerial 22 mm long. If the speed of the boat is 1.50 ms−1,1.50\,m{s^{ - 1}}, the magnitude of the induced emfemf in the wire of aerial is :
A 0.750.75 mVmV
B 0.500.50 mVmV
C 0.150.15 mVmV
D 11 mVmV
Correct Answer
Option C
Solution

Induced

emfemf
=vBHl=1.5×5×10−5×2= v{B_H}l = 1.5 \times 5 \times {10^{ - 5}} \times 2
=15×10−5=0.15 mV= 15 \times {10^{ - 5}} = 0.15\,mV
Q9
A circular loop of radius 0.30.3 cmcm lies center of the small loop is on the axis of the bigger loop. The distance between their centers is 1515 cm.cm. If a current of 2.02.0 AA flows through the smaller loop, than the flux linked with bigger loop is
A 9.1×10−11 9.1 \times {10^{ - 11}}\, weber
B 6×10−11 6 \times {10^{ - 11}}\, weber
C 3.3×10−11 3.3 \times {10^{ - 11}}\, weber
D 6.6×10−9 6.6 \times {10^{ - 9}}\, weber
Correct Answer
Option A
Solution

As we know, Magnetic flux,

ϕ=B.A\phi = B.A
μ0(2)(20×10−2)22[(0.2)2+(0.15)2]×π(0.3×10−2)2{{{\mu _0}\left( 2 \right){{\left( {20 \times {{10}^{ - 2}}} \right)}^2}} \over {2\left[ {{{\left( {0.2} \right)}^2} + {{\left( {0.15} \right)}^2}} \right]}} \times \pi {\left( {0.3 \times {{10}^{ - 2}}} \right)^2}

On solving

=9.216×10−11=9.2×10−11  = 9.216 \times {10^{ - 11}} = 9.2 \times {10^{ - 11}}\,\,

weber

Q10
Two coaxial solenoids of different radius carry current II in the same direction. F1→\overrightarrow {{F_1}} be the magnetic force on the inner solenoid due to the outer one and F2→\overrightarrow {{F_2}} be the magnetic force on the outer solenoid due to the inner one. Then :
A F1→\overrightarrow {{F_1}} is radially in wards and F2→=0\overrightarrow {{F_2}} = 0
B F1→\overrightarrow {{F_1}} is radially outwards and F2→=0\overrightarrow {{F_2}} = 0
C F1→=F2→=0\overrightarrow {{F_1}} = \overrightarrow {{F_2}} = 0
D F1→\overrightarrow {{F_1}} is radially inwards and F2→\overrightarrow {{F_2}} is radially outards
Correct Answer
Option C
Solution
F1→=F2→=0\mathop {F{}_1}\limits^ \to = \mathop {F{}_2}\limits^ \to = 0

because of action and reaction pair

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