Electromagnetic Induction Test

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Electromagnetic Induction Test - 18

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Question 1
1 / -0

A uniform magnetic field is restricted within a region of radius $$r$$. The magnetic field changes with time at a rate $$\dfrac{d\overrightarrow{B}}{dt}$$. Loop $$1$$ of radius $$R > r$$ enclose the region $$r$$ and loop $$2$$ of radius $$R$$ is outside the region of magnetic field as shown in the figure below. Then the $$e.m.f$$. generated is :

A
$$-\dfrac{d\overrightarrow{B}}{dt} \pi r^2$$ in loop $$1$$ and zero in loop $$2$$

B
Zero in loop $$1$$ and zero in loop $$2$$

C
$$-\dfrac{d\overrightarrow{B}}{dt} \pi r^2$$ in loop $$1$$ and $$-\dfrac{d\overrightarrow{B}}{dt} \pi r^2$$ in loop $$2$$

D
$$-\dfrac{d\overrightarrow{B}}{dt} \pi R^2$$ in loop $$1$$ and zero in loop $$2$$

Solution

Emf is induced in a coil due to change of flux through it.

In the loop 1, the flux passing through it is $$B\pi r^2$$

Hence emf induced=$$-\dfrac{d\phi}{dt}=-\dfrac{d\vec{B}}{dt}\pi r^2$$

Since the magnetic field does not change inside loop 2, no emf is induced in it.
Question 2
1 / -0

A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is

A

B

C

D

Solution

We know that electromagnetic e.m.f. induced,
$$e=\displaystyle -\frac{d\phi}{dt}$$

Initially, $$d\phi$$ will be positive(during first $$\displaystyle\frac{T}{4}$$ times period) then it becomes negative during the period from $$\displaystyle\frac{T}{4}$$ to $$\displaystyle\frac{T}{2}$$.

During the period $$\displaystyle\frac{T}{2}$$ to $$\displaystyle\frac{3T}{4}$$ it is again positive and in the last $$\displaystyle\frac{T}{4}$$ time it is negative.

Accordingly sign of emf produced will be changed. Figure $$(a)$$ fits exactly in this change pattern. So this figure represents the answer.
Question 3
1 / -0

Two solenoids of equal number of turns having their length and the radii in the same ratio $$1 : 2$$. The ratio of their self-inductance will be

A
$$1 : 2$$

B
$$2 : 1$$

C
$$1 : 1$$

D
$$1 : 4$$

Solution

Self inductance of a solenoid,
$$L = \dfrac{\mu_0 N^2 A}{l} = \dfrac{\mu_0 N^2 \pi r^2 }{l}$$
where $$l$$ is the length of the solenoid, N is the total number of turns of the solenoid and A is the area of cross-section of the solenoid.
$$\therefore \dfrac{L_1}{L_2} = \left( \dfrac{N_1}{N_2} \right)^2 \left( \dfrac{r_1}{r_2} \right)^2 \left( \dfrac{l_2}{l_1} \right)$$
Here, $$N_1 = N_2, \dfrac{l_1}{l_2} = \dfrac{1}{2} , \dfrac{r_1}{r_2} = \dfrac{1}{2}$$
$$\therefore \dfrac{L_1}{L_2} = \left( \dfrac{1}{2} \right)^2 \left( \dfrac{2}{1} \right) = \dfrac{1}{2}$$
Question 4
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In a coil of self inductance of $$5$$ henry, the rate of change of current is $$2$$ ampere per second, the e.m.f. induced in the coil is

A
$$5V$$

B
$$-5V$$

C
$$-10V$$

D
$$10V$$

Solution

e.m.f.$$=\displaystyle -L\frac{di}{dt}$$

$$=-5\times 2$$

$$=-10V$$
Question 5
1 / -0

Two concentric coils each of radius equal to $$2\pi $$cm are placed at right angles to each other. $$3$$ ampere and $$4$$ ampere are the currents flowing in each coil respectively. The magnetic induction in $$weber/m^2$$ at the centre of the coils will be $$\left(\mu_0=4\pi\times 10^{-7}wb/ A.m\right)$$

A
$$10^{-5}$$

B
$$12\times 10^{-5}$$

C
$$7\times 10^{-5}$$

D
$$5\times 10^{-5}$$

Solution

$$B_1=\displaystyle\frac{mu_0 i_1}{2(2\pi)}=\frac{\mu_0\times 3}{4\pi}$$

$$B_2=\displaystyle\frac{mu_0i_2}{2(2\pi)}=\frac{mu_0\times 4}{4\pi}$$

$$B=\sqrt{B^2_1+B^2_2}=\displaystyle \frac{mu_0}{4\pi}\cdot 5$$

$$\Rightarrow B=10^{-7}\times 5\times 10^2$$

$$\Rightarrow B=5\times 10^{-5}Wb/m^2$$
Question 6
1 / -0

Henry, the SI unit of inductance can be written as :

A
weber ampere$$^{-1}$$

B
volt second ampere$$^{-1}$$

C
joule ampere$$^{-1}$$

D
ohm s$$^{-1}$$

Solution

The SI unit of inductance is Henry,
$$\displaystyle L =-\dfrac{e}{\dfrac{di}{dt}}$$
SI unit $$= \displaystyle \dfrac{volt}{A} \times s$$
$$=volt \times second \times ampere^{-1}$$
Question 7
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Assertion (A): Whenever the magnetic flux linked with a closed coil changes there will be an induced emf as well as an induced current.
Reason (R): According to Faraday, the induced emf is inversely proportional to the rate of change of magnetic flux linked with a coil.

A
Both A and R are individually true and R is the correct explanation of A

B
Both A and R are individually true but R is not the correct explanation of A

C
A is true but R is false

D
Both A and R are false

Solution

$$EMF\quad =-\dfrac {d\phi }{ dt } \quad $$
So, EMF is directly proportional to the rate of change of magnetic flux linked with a coil. When closed coil is used current flows into it.
Question 8
1 / -0

The coefficient of self inductance and the coefficient of mutual inductance have

A
same units but different dimensions

B
different units but same dimensions

C
different units and different dimensions

D
same units and same dimensions

Solution

$$Emf=-L\dfrac { di }{ dt } $$

$$\phi =Mi$$

$$Emf\dfrac { d\phi }{ dt } =-M\dfrac { di }{ dt } $$

So, units and dimension of coeffiecent of self inductance and coefficent of mutual inductance are same.
Question 9
1 / -0

A varying current in a coil change from $$10A$$ to $$0$$ in $$0.5$$sec. If the average emf induced in the coil is $$220V$$, the self inductance of the coil is :

A
$$5H$$

B
$$6H$$

C
$$11H$$

D
$$12H$$

Solution

$$emf=L\dfrac { di }{ dt } $$

$$L=\dfrac { emf }{ (\dfrac { di }{ dt } ) } $$
$$L=\dfrac { 220 }{ (\dfrac { 10 }{ 0.5 }) } $$
$$=11H$$
Question 10
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Two conducting circular loops F and G are kept in a plane on either side of a straight current-carrying wire as shown in the figure below.
If the current in the wire decreases in magnitude, the induced current in the loops will be

A
clockwise in F and clockwise in G.

B
anti-clockwise in F and clockwise in G.

C
clockwise in F and anti-clockwise in G.

D
anti-clockwise in F and anti-clockwise in G.

Solution

The magnetic field above the wire is out of the plane. This flux is decreasing and should be compensated by the current in the loop F and so the current in loop F will be anti-clockwise. For loop G the situation is opposite