Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

A proton is moving with a velocity of $$3\times 10^7 m/s$$ in the direction of a uniform magnetic field of $$0.5 T$$. The force acting on proton is

A
$$2 N$$

B
$$4 N$$

C
$$6 N$$

D
zero

Solution

$$\overrightarrow{F}_{magnetic} = q(\overrightarrow{v} \times \overrightarrow{B})$$
Since $$\overrightarrow{v} \parallel \overrightarrow{B}$$
$$\overrightarrow{F}_{magnetic} =0$$
Question 2
1 / -0

Two parallel wires P and Q carry electric currents of $$10 A$$ and $$2 A$$ respectively in mutually opposite directions. The distance between the wires is $$10 cm$$. If the wire P is of infinite length and wire Q is $$2 m$$ long, then the force acting Q will be

A
$$4\times 10^{-5}N$$

B
$$8\times 10^{-5}N$$

C
$$6\times 10^{-5}N$$

D
$$ 14 \times 10^{-5}N$$

Solution

Force on a current carrying conductor is $$F=\int i (\vec{dl} \times \vec{B})$$
Magnetic field due to one wire at the location of the second wire ( other one ) is $$B_1= \dfrac{\mu_0 i_1}{2\pi d}$$
Force on the 2nd wire due to $$B_1$$ is $$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$$
Force per unit length for wire $$l_2$$ is $$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}$$
$$B_1$$ is in $$-\hat{k}$$ direction.
$$l_2$$ is in $$\hat{j}$$ direction.
Hence, $$F_{21}$$ will be in $$\hat{j} \times (\hat{k})=\hat{i}$$ direction.
$$ \therefore$$ it is an repulsive force.
$$F_{21} = \dfrac{2 \times 10^{-7} \times 10 \times 2 \times 2}{10 \times 10^{-2}}=8 \times 10^{-5}\:N$$
Question 3
1 / -0

The force between two parallel conductors, each of length $$50 m$$ and distant $$20 cm$$ apart, is $$1 N$$. If the current in one conductor is double than that in another one, then their values will respectively be :

A
$$100 A$$ and $$200 A$$

B
$$50 A$$ and $$400 A$$

C
$$10 A$$ and $$30 A$$

D
$$5 A$$ and $$25 A$$

Solution

Given that,
$$F = 1N$$,$$l = 50 m$$,$$r = 20 cm = 0.2 m$$,$$i_2 = 2i_1$$
The force between two long parallel conductors separated by distance r, is
$$F = \dfrac{\mu_0 (i_1)(i_2)l}{2 \pi r}$$
$$F = \dfrac{(4\pi \times 10^{-7}) (i_1)(2i_1)50}{2 \pi 0.2}$$
$$i_1^2 = \dfrac{2 \pi (0.2)}{(8\pi \times 10^{-7})50}$$
$$i_1^2 = \dfrac{0.2}{(200 \times 10^{-7})}$$
$$i_1^2 = 10000$$
$$i_1 = 100 A$$
and $$i_2 = 2i_1 = 200 A$$
Question 4
1 / -0

On passing electric current in two long straight conductors in mutually opposite directions, the magnetic force acting between them will be

A
attractive

B
repulsive

C
both attractive and repulsive

D
neither attractive nor repulsive

Solution

If electric current in two long straight conductors is in mutually opposite directions, the magnetic force acting between them will be repulsive.
Question 5
1 / -0

Two current-carrying parallel conductors are shown in the figure. The magnitude and nature of force acting between them per unit length will be :

A
$$8\times 10^{-8}N/m$$, attractive

B
$$3.2\times 10^{-5}N/m$$, repulsive

C
$$3.2\times 10^{-5}N/m$$, attractive

D
$$8\times 10^{-8}N/m$$, repulsive

Solution

Force on a current carrying conductor is $$F=\int i (\vec{dl} \times \vec{B})$$
Magnetic field due to one wire at the location of the second wire ( other one ) is $$B_1= \dfrac{\mu_0 i_1}{2\pi d}$$
Force on the 2nd wire due to $$B_1$$ is $$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$$
Force per unit length for wire $$l_2$$ is $$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}=\dfrac{\mu_0 i_1i_2}{2\pi d}=\dfrac{2 \times 10^{-7} \times 4 \times 4 }{10 \times 10^{-2}}=3.2 \times 10^{-5}\: N$$
where $$F_{21}$$ is the force on wire 2 due to wire 1 and the subscript 2 refers to 2nd wire.
$$B_1$$ is in $$-\hat{k}$$ direction.
$$l_2$$ is in $$-\hat{j}$$ direction.
Hence, $$F_{21}$$ will be in $$\hat{j} \times (-\hat{k})=-\hat{i}$$ direction.
Similarly, $$F_{12}$$ will be in $$\hat{i}$$ direction.
Thus, both the wires will attract each other.
Question 6
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A stream of electrons is projected horizontally to the right. A straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right, what will be the effect on the electron stream?

A
The electron stream will be pushed downwards

B
The electron stream will be pulled upwards

C
The electron stream will be retarded

D
The electron beam will be speeded up towards the right

Solution

The electron stream is equivalent to a current carrying conductor from right to left.
Two conductors carrying current in opposite directions will repel each other.
If the current carrying conductor is above the electron stream, the electron stream gets repelled and is pushed downwards.
If the current carrying conductor is below the electron stream, the electron stream gets repelled and is pushed upwards..
Question 7
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There is a magnetic field acting in a plane perpendicular to this sheet of paper, downward into the paper as shown in the figure. The particles in vacuum move in the plane of paper from left to right. The path indicated by the arrow could be traveled by :

A
proton

B
neutron

C
electron

D
a-particle

Solution

$$\vec{F} =q(\vec{v} \times \vec{B})= qvB( \hat{i} \times (-\hat{k}))=qvB \hat{j}$$
As seen from the diagram, the particle experiences force in $$-\hat{j}$$ direction.
$$ \Rightarrow$$ charge of the particle $$q$$ is negative.
Question 8
1 / -0

Current of 10 ampere and 2 ampere are passed through two parallel wires A and B, respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is 2 m, the force on the conductor B which is situated at 10 cm distance from A will be

A
$$8\times 10^{-5}N$$

B
$$4\times 10^{-5}N$$

C
$$8\pi \times 10^{-7}N$$

D
$$4\pi \times 10^{-7}N$$

Solution

Using Amper's law $$\oint \vec{B}.\vec{dl} = \mu_0 i_{enclosed}$$
$$ \therefore B = \dfrac{\mu_0 i_{enclosed}}{2\pi r}= \dfrac{4\pi \times 10^{-7} \times 10 }{2 \pi \times(10 \times 10 ^{-2})}= 2 \times 10^{-5} \:T$$
Force on the conductor $$B$$ is $$F= ilB = 2 \times 2 \times (2 \times 10 ^{-5})= 8 \times 10^{-5} \:N$$
Question 9
1 / -0

Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. the force of attraction per unit length between the two wires is

A
$$2\times 10^{-7}Nm^{-1}$$

B
$$2\times 10^{-8}Nm^{-1}$$

C
$$5\times 10^{-8}Nm^{-1}$$

D
$$10^{-7}Nm^{-1}$$

Solution

Force on a current carrying conductor is $$F=\int i (\vec{dl} \times \vec{B})$$
Magnetic field due to one wire at the location of the second wire ( other one ) is $$B_1= \dfrac{\mu_0 i_1}{2\pi d}$$
Force on the 2nd wire due to $$B_1$$ is $$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$$
Force per unit length for wire $$l_2$$ is $$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}=\frac{\mu_0 i_1i_2}{2\pi d}=\dfrac{2 \times 10^{-7} \times 1 \times 1 }{1}=2 \times 10^{-7} \: N$$
where $$F_{21}$$ is the force on wire 2 due to wire 1 and the subscript 2 refers to 2nd wire.
$$B_1$$ is in $$-\hat{k}$$ direction.
$$l_2$$ is in $$-\hat{j}$$ direction.
Hence, $$F_{21}$$ will be in $$\hat{j} \times (-\hat{k})=-\hat{i}$$ direction.
Similarly, $$F_{12}$$ will be in $$\hat{i}$$ direction.
Thus, both the wires will attract each other.
Question 10
1 / -0

Directions For Questions

You are asked to do an experiment to study the effect of magnetic field on charged particle. You take two long wires having resistance $$10\Omega$$ and $$25\Omega$$. Separated them by $$5 cm$$ and keep them parallel. The two are connected to a battery of $$100 V$$ as shown in Figure. The battery branch is kept quite far away from the two conductors. A proton is allowed to enter the plane of the wires directed towards the upper wire with a velocity 650 $$km /s$$ exactly in the middle of the wire.

...view full instructions

The initial acceleration of the proton is

A
$$2.9\times 10^8 ms^{-2}$$

B
$$3.31\times 10^2 ms^{-2}$$

C
$$3.12\times 10^8 ms^{-2}$$

D
none of these

Solution

Current $$I_1=\dfrac{100}{10}A$$ = $$10A$$ Current $$I_2=\dfrac{100}{25}A$$ = $$4A$$
Force on moving proton due to magnetic field is $$F=qvB$$
$$F=qv\left[ \cfrac {\mu_0I_1}{2\pi d}-\cfrac {\mu_0I_2}{2\pi d}\right]=\cfrac{qv\mu_0}{2\pi d} \left[I_1- I_2 \right]$$
$$ma=\cfrac{qv\mu_0}{2\pi d} \left[I_1-I_2 \right]$$
$$a=\dfrac {1.6\times 10^{-19}\times 650\times 10^3[10-4]\times 4\pi \times 10^{-7}}{2\pi \times 2.5\times 10^{-2}\times 1.6\times
10^{-27}}$$
$$a =3.12\times 10^8ms^{-2}$$

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Moving Charges and Magnetism Test - 42

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Moving Charges and Magnetism Test - 42

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Question 1

$$2 N$$

$$4 N$$

$$6 N$$

zero

Solution

Question 2

$$4\times 10^{-5}N$$

$$8\times 10^{-5}N$$

$$6\times 10^{-5}N$$

$$ 14 \times 10^{-5}N$$

Solution

Question 3

$$100 A$$ and $$200 A$$

$$50 A$$ and $$400 A$$

$$10 A$$ and $$30 A$$

$$5 A$$ and $$25 A$$

Solution

Question 4

attractive

repulsive

both attractive and repulsive

neither attractive nor repulsive

Solution

Question 5

$$8\times 10^{-8}N/m$$, attractive

$$3.2\times 10^{-5}N/m$$, repulsive

$$3.2\times 10^{-5}N/m$$, attractive

$$8\times 10^{-8}N/m$$, repulsive

Solution

Question 6

The electron stream will be pushed downwards

The electron stream will be pulled upwards

The electron stream will be retarded

The electron beam will be speeded up towards the right

Solution

Question 7

proton

neutron

electron

a-particle

Solution

Question 8

$$8\times 10^{-5}N$$

$$4\times 10^{-5}N$$

$$8\pi \times 10^{-7}N$$

$$4\pi \times 10^{-7}N$$

Solution

Question 9

$$2\times 10^{-7}Nm^{-1}$$

$$2\times 10^{-8}Nm^{-1}$$

$$5\times 10^{-8}Nm^{-1}$$

$$10^{-7}Nm^{-1}$$

Solution

Question 10

$$2.9\times 10^8 ms^{-2}$$

$$3.31\times 10^2 ms^{-2}$$

$$3.12\times 10^8 ms^{-2}$$

none of these

Solution

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