Electromagnetic Induction Test

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Electromagnetic Induction Test - 24

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Question 1
1 / -0

To measure the field $$B$$ between the poles of an electromagnet, a small test loop of area $$1$$$$cm^{2}$$, resistance $$10$$$$\Omega $$ and $$20$$ turns is pulled out of it. A galvanometer shows that a total charge of $$2 \mu C $$ passed through the loop. The value of $$B$$ is

A
$$0.001 T$$

B
$$0.01 T$$

C
$$0.1 T$$

D
$$1.0 T$$

Solution

$$emf=\dfrac { BAn }{ t } $$
$$i=\dfrac { emf }{ R } =\dfrac { BAn }{ tR } $$
$$q=it=\dfrac { BAn }{ tR } =\dfrac { BAn }{ R } $$
$$\therefore 2\times { 10 }^{ -6 }=\dfrac { B\times { 10 }^{ -4 }\times 20 }{ 10 } $$
$$B={ 10 }^{ -2 }T$$
$$B= 0.01T$$
Question 2
1 / -0

An average emf of $$32 V$$ is induced in a coil in which the current drops from $$10 A$$ to $$2 A$$ in $$0.1 s$$. The inductance of the coil is:

A
$$0.32 H$$

B
$$0.4 H$$

C
$$4 H$$

D
$$0.04 H$$

Solution

$$emf=L\dfrac { di }{ dt } $$
$$emf=L\dfrac { \triangle i }{ \triangle t } $$

So, $$L=emf\times \dfrac { \triangle t }{ \triangle i } $$
= $$32\times \dfrac { 0.1 }{ 8 } $$
$$= 0.4H$$
Question 3
1 / -0

A wire moves with a velocity $$v$$ through a magnetic field and experiences an induced charge separation as shown in the figure. Then the direction of the magnetic field is

A
into the page.

B
out of the page.

C
towards the bottom of the page.

D
towards the top of the page.

Solution

Force on change is given by $$q(V\times B)$$
So, $$F = q (v\times B)$$
So, force direction is in the direction of $$V \times B$$
So, by using right hand rule for determing the direction of $$V\times B$$, we get direction of B into the plane of page.
Question 4
1 / -0

A train is moving towards north with a speed of $$180$$ kilometers per hour. If the vertical component of the earths magnetic filed is $$0.2 \times 10^{- 4}$$ $$T$$, the emf induced in the axle of length $$1.5$$ $$m$$ is

A
$$1.5 mV$$

B
$$15 mV$$

C
$$54 mV$$

D
$$5.4 mV$$

Solution

$$emf = B l V$$
= $$0.2\times { 10 }^{ -4 }\times 1.5\times 180\times \dfrac { 1000 }{ 3600 } $$
= $$1.5\times { 10 }^{ -3 }V$$
= $$1.5$$ $$mV$$
Question 5
1 / -0

When rate of change of current in a circuit is unity, the induced emf is equal to

A
total flux linked with the coil.

B
induced charge.

C
number of turns in the circle.

D
coefficient of self induction.

Solution

Induced emf is equal to
$$emf =-\dfrac { d(N\emptyset ) }{ dt } = \dfrac { -d\left( Li \right) }{ dt } =-L\dfrac { di }{ dt } $$

So, when rate of change of current is equal to unity, emf is equal to coefficent of self inductance.
Question 6
1 / -0

An electric potential difference will be induced between the ends of the conductor shown in the figure, if the conductor moves in the direction shown by :

A
P

B
R

C
L

D
M

Solution

Hint:

Fleming’s Right Hand Rule gives the direction of current in the conductor whenever it is moved in any magnetic field. It states that if we stretch our right hand fingers perpendicular to each other as shown in fig, then if fore finger is in the direction of Magnetic field and thumb is in the direction of motion of conductor then middle finger gives the direction of current in the conductor.

Correct Option is D.

Explanation for the correct answer:
$$\bullet$$ For induction of emf in the conductor, the motion of conductor should be perpendicular to the direction of magnetic field. Thus, Option B and C are eliminated.
$$\bullet$$ Now if we move conductor in P direction, the motion will be perpendicular to the direction of magnetic field but then the charge separation in the conductor will take place perpendicular to the length of the conductor. Thus, due to less space for charge separation emf can not be induced in the conductor. Thus, Option A is eliminated.
$$\bullet$$ If conductor is moved in direction M then the charge separation in the conductor takes place across its length, there will be enough space for charge separation. And thus emf will be induced across the conductor.
Question 7
1 / -0

A coil has $$100$$ turns. When a current of $$5A$$ produces a magnetic flux of $$1\mu Wb$$, the coefficient of self induction is :

A
$$10\mu H $$

B
$$20\mu H $$

C
$$30\mu H $$

D
$$40\mu H $$

Solution

$$N\phi =Li$$
$$L=\dfrac { N\phi }{ i } $$
=$$\dfrac { 100\times 1\times { 10 }^{ -6 } }{ 5 } $$
=$$20\times { 10 }^{ -6 }H$$
=$$20\mu H$$
Question 8
1 / -0

The rate of change of current needed to induce an emf of $$8 V$$ in $$0.1 H$$ coil is

A
$$0.8 A/s$$

B
$$0.125 A/s$$

C
$$80 A/s$$

D
$$8 A/s$$

Solution

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

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

$$=\dfrac { 8 }{ 0.1 } $$

$$=80A/s $$
Question 9
1 / -0

Assertion : The induced emf and current will be same in two identical loops of copper and aluminium, when rotated with same speed in the same magnetic field.
Reason : Mutual induction does not depend on the orientation of the coils.

A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

C
Assertion is true but Reason is false

D
Assertion and Reason both are false

Solution

A and R both are false.
A: the emf will be same but current will different as resistance will be different of Cu and Al.
R: mutual inductance depends on orientation of coil, size, shape, no. of turns and relative position.
Question 10
1 / -0

In an inductance coil the current increases from zero to $$6A$$ in $$0.3$$ second by which an induced e.m.f. of $$60V$$ is produced in it. The value of coefficient of self-induction of coil is :

A
$$1H$$

B
$$0.5H$$

C
$$2H$$

D
$$3H$$

Solution

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

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

$$=\dfrac { 60 }{ (\dfrac { 6 }{ 0.3 } ) } $$
$$= 3H$$

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Electromagnetic Induction Test - 24

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Electromagnetic Induction Test - 24

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Question 1

$$0.001 T$$

$$0.01 T$$

$$0.1 T$$

$$1.0 T$$

Solution

Question 2

$$0.32 H$$

$$0.4 H$$

$$4 H$$

$$0.04 H$$

Solution

Question 3

into the page.

out of the page.

towards the bottom of the page.

towards the top of the page.

Solution

Question 4

$$1.5 mV$$

$$15 mV$$

$$54 mV$$

$$5.4 mV$$

Solution

Question 5

total flux linked with the coil.

induced charge.

number of turns in the circle.

coefficient of self induction.

Solution

Question 6

P

R

L

M

Solution

Question 7

$$10\mu H $$

$$20\mu H $$

$$30\mu H $$

$$40\mu H $$

Solution

Question 8

$$0.8 A/s$$

$$0.125 A/s$$

$$80 A/s$$

$$8 A/s$$

Solution

Question 9

Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Assertion is true but Reason is false

Assertion and Reason both are false

Solution

Question 10

$$1H$$

$$0.5H$$

$$2H$$

$$3H$$

Solution

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