Electromagnetic Induction Test

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Electromagnetic Induction Test - 38

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Question 1
1 / -0

The current $$i$$ in an induction coil varies with time $$t$$ according to the graph shown in the figure. Which of the following graphs shows the induced emf ($$\varepsilon $$) in the coil with time?

A

B

C

D

Solution

When the current is constant, the induced emf is zero as $$emf=-\dfrac { Ldi }{ dt } $$ and when the current decreases at constant rate the emf is a positive constant and when it increase at constant rate it is a negative constant.
Question 2
1 / -0

In figure, if the current $$i$$ decreases at a rate $$\alpha$$, then $${V}_{A}-{V}_{B}$$ is

A
zero

B
$$-\alpha L$$

C
$$\alpha L$$

D
No relation exists

Solution

For an inductor
$$ \Delta V = L \cfrac{di}{dt}$$
Given
$$\cfrac{di}{dt}= - \alpha$$
$$V_A-V_B = - \alpha L$$
Question 3
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In the figure shown, a T-shaped conductor moves with constant angular velocity $$\omega$$ in a plane perpendicular, to uniform magnetic field $$\vec { B } $$. The potential difference $${V}_{A}-{V}_{B}$$ is

A
zero

B
$$\cfrac { 1 }{ 2 } B\omega { l }^{ 2 }$$

C
$$2B\omega { l }^{ 2 }$$

D
$$B\omega { l }^{ 2 }$$

Solution

Since potential difference in a moving conductor is produced in a direction perpendicular to the direction of velocity, potential difference is not produced along the length of AB.
Question 4
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If the instantaneous magnetic flux and induced emf produced in a coil is $$\phi$$ and $$E$$ respectively, then according to Faraday's law of electro magnetic induction:

A
$$E$$ must be zero if $$\phi=0$$ and changing

B
$$E\ne 0$$ if $$\phi=0$$

C
$$E\ne 0$$ if $$\phi$$ is changing.

D
$$E=0$$ then $$\phi$$ must be zero

Solution

Farady's law states that time varying magnetic flux can induce an e.m.f.
E = Electric field, Induced
$$E = - \cfrac{d \phi}{dt}$$

$$E = - \cfrac{d \phi}{dt}$$
$$E =0$$ only if $$ \cfrac{d \phi}{dt} =0$$
If $$E=0$$, then $$\phi = const.$$
$$E \ne 0$$, then $$ \phi $$ is changing
Question 5
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In figure, there is a conducting ring having resistance $$R$$ placed in the plane of paper in a uniform magnetic field $${B}_{0}$$. If the ring is rotating in the plane of paper about an axis passing through point $$O$$ and perpendicular to the plane of paper with constant angular speed $$\omega$$ in clockwise direction, then

A
point $$O$$ will be at higher potential than $$A$$

B
the potential of point $$B$$ and $$C$$ will different

C
the current in the ring will be zero

D
the current in the ring will be $$2{B}_{0}\omega {r}^{2}/R$$

E
answer required

Solution

The ring can be seen as two rod OCA and OBA in parallel.

The motional emf will be induced between O and A = $$\dfrac{1}{2} B (2r)^2 \omega = 2Br^2 \omega $$ with A positibe wrt O

In both the paths from O - C - A ; and from O - B - A , potential increases as one moves from point O to point A.
Point B and C are at same potential due to symmetry.
As there is no return path, no current flows even though a potential is developed between points O and A
Question 6
1 / -0

A wooden stick of length $$3l$$ is rotated about an end with constant angular velocity $$\omega$$ in a uniform magnetic field $$B$$ perpendicular to the plane of motion. If the upper one-third of its length is coated with copper, the potential difference across the whole length of the stick is

A
$$\dfrac{9B\omega {l}^{2}}{2}$$

B
$$\dfrac{4B\omega {l}^{2}}{2}$$

C
$$\dfrac{5B\omega {l}^{2}}{2}$$

D
$$\dfrac{B\omega {l}^{2}}{2}$$

Solution

Force on one electron in the rod, at a point r form center, due to the motion = $$F= q \vec v \times \vec B = q_e r \omega B $$

total emf = $$ \int_{2l}^{3l}{\dfrac{F}{q_e} dr}= \int_{2l}{3l}{ r \omega B dr} $$

$$= B \omega (\dfrac{9l^2}{2} - \dfrac{4l^2}{2} ) = B \omega \dfrac{ 5 l^2}{2} $$
Question 7
1 / -0

A horizontal straight conductor when placed along south-north direction falls under gravity; there is

A
an induced current from south-to-north direction

B
an induced emf current from north-to-south direction

C
no induced emf along the length of the conductor

D
an induced emf along the length of the conductor

Solution

$$\bullet$$We Know induced emf $$\vec E=\vec B.(\vec l \times \vec v)$$
$$\bullet$$The earth's magnetic field lines are along the north south direction which is parallel to the length of the moving rod.
Therefore, no emf is induced along the length in the rod.

$$Answer:$$

Hence, option C is the correct answer.
Question 8
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A metal disc of radius $$a$$ rotates with a constant angular velocity $$\omega$$ about its axis. The potential difference between the center and the rim of the disc is ($$m=$$ mass of electron, $$e=$$ charge on electron)

A
$$\cfrac{m{\omega}^{2}{a}^{2}}{e}$$

B
$$\cfrac { 1 }{ 2 } \cfrac{m{\omega}^{2}{a}^{2}}{e}$$

C
$$\cfrac{e{\omega}^{2}{a}^{2}}{2m}$$

D
$$\cfrac{e{\omega}^{2}{a}^{2}}{m}$$

Solution

For second law of Motion,
$$F=ma$$
$$eE=mw^2r$$
$$E=\dfrac{mw^2r}{e}$$
The potential difference $$V=\int E.dr$$
$$V=\dfrac{\int_{0}^{a}mw^2r}{e}=\dfrac{mw^2a^2}{2e}$$
Question 9
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Two identical cycle wheels (geometrically) have different number of spokes connected from center to rim. One is having $$20$$ spokes and the other having only $$10$$ (the rim and the spokes are resistanceless). One resistance of value $$R$$ is connected between center and rim. The current in $$R$$ will be

A
double in the first wheel than in the second wheel

B
four times in the first wheel than in the second wheel

C
will be double in the second wheel than that of the first wheel

D
will be equal in both these wheels

E
answer required

Solution
Question 10
1 / -0

A uniform magnetic field exists in a region given by $$\vec { B } =3\hat { i } +4\hat { j } +5\hat { k } $$. A rod of length $$5m$$ along y-axis moves with a constant speed of $$1m$$ $${s}^{-1}$$ along x axis. Then the induced emf in the rod will be

A
$$0$$

B
$$25V$$

C
$$20V$$

D
$$15V$$

E
none

Solution

$$ Given, B = 3 + 4 + 5k $$
$$V = 1 m/s (As \ the \ rod \ moves \ in \

X-Direction)$$
$$Length \ of \ rod = 5 m$$
$$Since, e = -d\phi /dt \ and \ \phi = B.A $$

$$So, e = -d(B.A )/dt$$

$$ e = d(B.A)/dt   ( Negative \ sign \ is \

neglected \ as \ it \ only \ represents \ the \

direction)$$

$$ e = B.d(A)/dt   ( Magnetic \ Field \ is \ constant

)$$

$$ = B.d(l\times b)/dt ( Where \ l = Length \ and \ b = Breadth

)$$

$$ = l B. d(b)/dt ---- (1) $$

As $$db/dt$$ is nothing but the velocity with which the rod is moving.
$$So, db/dt = V $$
Putting the above value of $$db/dt$$ in equation (1)
We have, $$e = l B.V $$
But, in the above equation it is assumed that the direction of B, L and V are mutually perpendicular.
So, Only that component   of B that is in mutual perpendicular direction of L and V will be taken into account.
$$ So, e = l B.V   = 5 \times 5 \times 1 $$
    $$ e = 25 V$$