38: Electromagnetic Induction : Neet-xii-physics

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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 7

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#21-a
Find the average emf induced in the coil.
Ans : The emf induced in the coil is given by
`` e=-\frac{N∆\varphi }{∆t}=\frac{N(\stackrel{\to }{{B}_{f}}.{\stackrel{\to }{A}}_{f}-\stackrel{\to }{{B}_{i}}.{\stackrel{\to }{A}}_{i})}{T}``
`` =\frac{NB.A(\,\mathrm{\,cos\,}{0}^{\,\mathrm{\,o\,}}-\,\mathrm{\,cos\,}60°)}{T}``
`` =\frac{50\times 2\times {10}^{-1}\times \,\mathrm{\,\pi \,}(0.02{)}^{2}}{2\times 0.1}``
`` =5\times 4\times {10}^{-5}\times \,\mathrm{\,\pi \,}``
`` =2\,\mathrm{\,\pi \,}\times {10}^{-2}\,\mathrm{\,V\,}=6.28\times {10}^{-3}\,\mathrm{\,V\,}``

#21-b
If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 â„¦, calculate the net charge crossing a cross-section of the wire of the coil.
Ans : The current in the coil is given by
`` i=\frac{e}{R}=\frac{6.28\times {10}^{-3}}{4}``
`` =1.57\times {10}^{-3}\,\mathrm{\,A\,}``
The net charge passing through the cross section of the wire is given by
`` Q=it=1.57\times {10}^{-3}\times {10}^{-1}``
`` =1.57\times {10}^{-4}\,\mathrm{\,C\,}``
Page No 307:

Qstn #22
A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 × 10^-4 T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm² and its resistance is 4.0 â„¦. Find
Ans : Given:
Number of turns in the coil, n = 100 turns
Magnetic field, B = 4 × 10^-4
Area of the loop, A = 25 cm² = 25 × 10^-4 m²

#22-a
the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field,
Ans : When the coil is perpendicular to the field:
ϕ₁ = nBA
When the coil goes through the half turn:
ϕ₂ = nBA cos 180° = -nBA
∴ Δϕ = 2nBA
When the coil undergoes 300 revolutions in 1 minute, the angle swept by the coil is
300 × 2π rad/min = 10π rad/s
10π rad is swept in 1 s.
π rad is swept in `` \left(\frac{1}{10\,\mathrm{\,\pi \,}}\right)\,\mathrm{\,\pi \,}=\frac{1}{10}\,\mathrm{\,s\,}``
`` e=\frac{d\,\mathrm{\,\varphi \,}}{dt}=\frac{2nBA}{dt}``
`` =\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{1/10}``
`` =2\times {10}^{-3}\,\mathrm{\,V\,}``

#22-b
the average emf in a full turn and
Ans : ϕ₁ = nBA, ϕ₂ = nBA (θ = 360°)
Δϕ = 0, thus emf induced will be zero.

#22-c
the net charge displaced in part
(a).
Ans : The current flowing in the coil is given by
`` i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}``
= 0.5 × 10^-3 = 5 × 10^-4 A
Hence, the net charge is given by
Q = idt = 5 × 10^-4× `` \frac{1}{10}``
= 5 × 10^-5 C
Page No 307:

Qstn #23
A coil of radius 10 cm and resistance 40 â„¦ has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer if the horizontal component of the earth’s magnetic field is B_H = 3.0 × 10^-5 T.
Ans : Given:
Radius of the coil, r = 10 cm = 0.1 m
Resistance of the coil, R = 40 Ω
Number of turns in the coil, N = 1000
Angle of rotation, θ = 180°
Horizontal component of Earth's magnetic field, B_H = 3 × 10^-5 T
Magnetic flux, ϕ = NBA cos 180°
⇒ ϕ = -NBA
= -1000 × 3 × 10^-5 × π × 1 × 1 × 10^-2
= 3π × 10^-4 Wb
dϕ = 2NBA = 6π × 10^-4 Wb
`` e=\frac{d\,\mathrm{\,\varphi \,}}{dt}=\frac{6\pi \times {10}^{-4}}{dt}`` V
Thus, the current flowing in the coil and the total charge are:
`` i=\frac{e}{R}=\frac{6\,\mathrm{\,\pi \,}\times {10}^{-4}}{40dt}=\frac{4.71\times {10}^{-5}}{dt}``
`` Q=\frac{4.71\times {10}^{-5}\times dt}{dt}``
`` =4.71\times {10}^{-5}\,\mathrm{\,C\,}``
Page No 307:

Qstn #24
A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.010 T exists in a direction perpendicular to the axis of rotation. Find
Ans : Given,
Radius of the circular coil, R = 5.0 cm
Angular speed of circular coil, `` \omega `` = 80 revolutions/minute
Magnetic field acting perpendicular to the axis of rotation, B = 0.010 T
The emf induced in the coil `` \left(e\right)`` is given by,
`` e=\frac{d\varphi }{dt}``
`` \Rightarrow e=\frac{dB.A\,\mathrm{\,cos\,}\theta }{dt}``
`` \Rightarrow e=-BA\,\mathrm{\,sin\,}\theta \frac{d\theta }{dt}``
`` \Rightarrow ``e = -BAωsinθ
(`` \frac{d\theta }{dt}=\omega `` = the rate of change of angle between the arc vector and B)

#24-a
the maximum emf induced,
Ans : For maximum emf, sinθ = 1
`` \therefore `` e = BAω
`` \Rightarrow ``e = 0.010 × 25 × 10^-4 × 80 × `` \frac{2\,\mathrm{\,\pi \,}\times \,\mathrm{\,\pi \,}}{60}``
`` \Rightarrow ``e = 0.66 × 10^-3 = 6.66 × 10^-4 V

#24-b
the average emf induced in the coil over a long period and
Ans : The direction of the induced emf changes every instant. Thus, the average emf becomes zero.

#24-c
the average of the squares of emf induced over a long period.
Ans : The emf induced in the coil is e = -BAωsinθ = -BAωsin ωt
The average of the squares of emf induced is given by
`` {{e}_{\,\mathrm{\,av\,}}}^{2}=\frac{{\int }_{0}^{T}{B}^{2}{A}^{2}{\omega }^{2}{\,\mathrm{\,sin\,}}^{2}\omega t\,\mathrm{\,d\,}t}{{\int }_{0}^{T}\,\mathrm{\,d\,}t}``
`` \Rightarrow {{e}_{\,\mathrm{\,av\,}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega }^{2}{\int }_{0}^{T}{\,\mathrm{\,sin\,}}^{2}\omega t\,\mathrm{\,d\,}t}{{\int }_{0}^{T}\,\mathrm{\,d\,}t}``
`` \Rightarrow {{e}_{\,\mathrm{\,av\,}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega }^{2}{\int }_{0}^{T}\left(1-\,\mathrm{\,cos\,}2\omega t\right)\,\mathrm{\,d\,}t}{2T}``
`` \Rightarrow {{e}_{\,\mathrm{\,av\,}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega }^{2}}{2T}{\left[t-\frac{\,\mathrm{\,sin\,}2\omega t}{2\omega }\right]}_{0}^{T}``
`` \Rightarrow {{e}_{\,\mathrm{\,av\,}}}^{2}=\frac{{B}^{2}{A}^{2}{\omega }^{2}}{2T}\left[T-\frac{\,\mathrm{\,sin\,}4\pi -\,\mathrm{\,sin\,}0}{2\omega }\right]=\frac{{B}^{2}{A}^{2}{\omega }^{2}}{2}``
`` \Rightarrow {{e}_{\,\mathrm{\,av\,}}}^{2}=\frac{(6.66\times {10}^{-4}{)}^{2}}{2}=22.1778\times {10}^{-8}{\,\mathrm{\,V\,}}^{2}\left[\because BA\omega =6.66\times {10}^{-4}\,\mathrm{\,V\,}\right]``
`` \Rightarrow {{e}_{\,\mathrm{\,av\,}}}^{2}=2.2\times {10}^{-7}{\,\mathrm{\,V\,}}^{2}``
Page No 308:

Qstn #25
Suppose the ends of the coil in the previous problem are connected to a resistance of 100 â„¦. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.
Ans : Given:
T = 1 minute
Heat produced in the circuit is calculated using the following relation:
`` H=\underset{0}{\overset{T}{\int }}{i}^{2}Rdt``
`` \Rightarrow H=\underset{0}{\overset{1\,\mathrm{\,min\,}}{\int }}\frac{{B}^{2}{A}^{2}{\omega }^{2}}{{R}^{2}}\,\mathrm{\,sin\,}\left(\omega t\right)Rdt``
`` =\frac{{B}^{2}{A}^{2}{\omega }^{2}}{2R}.\underset{0}{\overset{1\,\mathrm{\,min\,}}{\int }}\left(1-\,\mathrm{\,cos\,}2\,\mathrm{\,\omega \,}t\right)dt``
`` =\frac{{B}^{2}{A}^{2}{\omega }^{2}}{2R}{\left(1-\frac{\,\mathrm{\,sin\,}2\omega t}{2\omega }\right)}_{0}^{1\,\mathrm{\,min\,}}``
`` =\frac{{B}^{2}{A}^{2}{\omega }^{2}}{2R}\left(60-\frac{\,\mathrm{\,sin\,}2\times 80\times 2\,\mathrm{\,\pi \,}/60\times 60}{2\times 80\times 2\,\mathrm{\,\pi \,}/60}\right)``
`` =\frac{60}{2R}\times {\pi }^{2}{r}^{4}\times {B}^{2}\times {\left(80\times \frac{2\pi }{60}\right)}^{2}``
`` =\frac{60}{200}\times 10\times \frac{64}{9}\times 10\times 625\times {10}^{-8}\times {10}^{-4}``
`` =\frac{625\times 6\times 64}{9\times 2}\times {10}^{-11}=1.33\times {10}^{-7}\,\mathrm{\,J\,}``
Page No 308:

Qstn #26
Figure
Figure shows a circular wheel of radius 10.0 cm whose upper half, shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic field B of magnitude 2.00 × 10^-4 T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period.
Ans : Magnetic flux through the wheel (initially):
`` {\varphi }_{1}=BA=\frac{2\times {10}^{-4}\times \,\mathrm{\,\pi \,}{\left(0.1\right)}^{2}}{2}``
`` =\,\mathrm{\,\pi \,}\times {10}^{-6}\,\mathrm{\,Wb\,}``
As the wheel rotates, the wooden (non-metal) part of the wheel comes inside the magnetic field and the iron part of the wheel comes outside the magnetic field. Thus, the magnetic flux through the wheel becomes zero.
i.e. `` {\varphi }_{2}=0``
dt = 2 s
The average emf induced in the wheel is given by
`` e=-\frac{d\,\mathrm{\,\varphi \,}}{dt}``
`` =-\left(\frac{{\varphi }_{2}-{\varphi }_{1}}{dt}\right)``
`` =\frac{\,\mathrm{\,\pi \,}\times {10}^{-6}}{2}``
`` =1.57\times {10}^{-6}\,\mathrm{\,V\,}``
Page No 308:

Qstn #27
A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm s^-1 perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion.
Ans : Given:
Length of the rod, l = 20 cm = 0.2 m
Velocity of the rod, v = 10 cm/s = 0.1 m/s
Magnetic field, B = 0.10 T

#27-a
Find the average magnetic force on the free electrons of the rod.
Ans : The force on a charged particle moving with velocity v in a magnetic field is given by
`` \stackrel{\to }{F}=q\left(\stackrel{\to }{v}\times \stackrel{\to }{B}\right)``
F = qvB sin θ
Here,
θ = 90^o
Now,
F = (1.6 × 10^-19) × (1 × 10^-1) × (1 × 10^-1)
= 1.6 × 10^-21 N