Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

A charged particle, having charge $$q$$ accelerated through a potential difference $$V$$ enters a perpendicular magnetic field in which it experiences a force $$F$$. If $$V$$ is increased to $$5 V$$,the particle will experience a force

A
$$F$$

B
$$5F$$

C
$$0.2F$$

D
$$\sqrt{5}F$$

Solution

$$\dfrac {1}{2}mv^2=qV$$
So $$ v=\sqrt {\dfrac {2qV}{m}}$$
$$F_1=qvB=q\sqrt{ \dfrac {2qV}{m}} B$$
$$F_2=q\sqrt {\dfrac {2q\times 5V}{m}} B$$
$$F_2=\sqrt 5 F_1$$
Question 2
1 / -0

A particle of charge $$q$$ and mass $$m$$ moving with a velocity $$v$$ along $$x-$$axis enters the region $$x > 0$$ with uniform magnetic field $$B$$ along the $$\hat k$$ direction. The particle will penetrate in this region in the $$x-$$direction up to a distance $$d$$ equal to

A
$$Zero$$

B
$$\dfrac{mv}{qB}$$

C
$$\dfrac{2mv}{qB}$$

D
$$Infinity$$

Solution

Lorentz force $$F=q(\vec{v}\times\vec{B})=q(v\hat{i}\times B\hat{k}) =-qvB\hat{j}$$
The particle move in a circle, the plane of the circle is perpendicular to the direction of magnetic field.
Centripetal force is provided by the Lorentz force.
$$qvB= \dfrac{mv^2}{r}$$
$$\Rightarrow r =\dfrac{mv}{qB}$$
Question 3
1 / -0

If the distance as well as the current in each of two parallel current carrying wires is doubled, the force per unit length acting between them becomes

A
doubled

B
remains same

C
quadrupled

D
halved

Solution

Force per unit length $$(F)$$ between two current carrying wires is
$$F=\dfrac { \mu_o{ I }_{ 1 }{ I }_{ 2 } }{ 2A\pi }$$ $$N$$
When both $$I_{1}$$, $$I_{2}$$ and $$r$$ are doubled then
$$\Rightarrow F'=$$ $$\dfrac { \mu_o(2{ I }_{ 1 })(2{ I }_{ 2 }) }{ 2A(2{ \pi }_{ 1 }) } = 2F$$
Question 4
1 / -0

The distance between the wires of electric mains is 12cm. These wires exprience 4 mgwt per unit length. The value of current flowing in each wire will be if they carry current in same direction

A
4.85A

B
zero

C
$$4.85\times 10^{-2}A$$

D
$$85\times 10^{-4}A$$

Solution

$$\dfrac {df}{dl}=\dfrac {\mu_0 i_1i_2}{2\pi r}$$
$$4=\dfrac {2\times 10^{-7}\times i^2}{12\times 10^{-2}}$$
$$i=4\cdot 85A$$
Question 5
1 / -0

The distance between two parallel wires carrying current of $$1A$$ is $$1m$$. The force per unit length between the conductors is

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

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

C
$$4\times 10^{7}Nm^{-1}$$

D
$$4\times 10^{-7}Nm^{-1}$$

Solution

$$F=\dfrac { \mu_0 }{ 2Ad } { i }_{ 1 }{ i }_{ 2 }$$
$$=\dfrac { 2\times { 10 }^{ -7 }\times 1\times 1 }{ 1 } =\quad 2\times 10^{ -7 }{ N }/{ m }$$
Question 6
1 / -0

In an electric motor, wires carrying a current of $$5A$$ are placed at right angles to a magnetic field of induction $$0.8 T$$. If each wire has length of $$20cm$$, then the force acting on each wire is :

A
$$0.2N$$

B
$$0.4N$$

C
$$0.6N$$

D
$$0.8N$$

Solution

$$F= Bil
= 0.8 \times 5 \times 20 \times 10^{-2}
= 0.8N $$
Question 7
1 / -0

Two parallel conductors carrying 5A each, repel with a force of 0.25 N $$m^{-1}$$ . The distance between them is

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

B
$$3\times 10^{-5}m$$

C
$$2\times 10^{-5}m$$

D
$$1\times 10^{-5}m$$

Solution

$$\dfrac {df}{dl}=\dfrac {\mu_0 i_1i_2}{2\pi r}$$
$$0\cdot 25=\dfrac {2\times 10^{-7}\times 5\times 5}{r}$$
$$r=2\times 10^{-5}m$$
Question 8
1 / -0

Two long straight wires carry currents 5 A and 10 A in the same direction are separated by 20 cm. The force between wires is

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

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

C
$$2\times 10^{-5}N/m$$ attractive

D
$$2\times 10^{-5}N/m$$ repulsive

Solution

$$\displaystyle F=\frac { Mo }{ 2A } \times \frac { { i }_{ 1 }{ i }_{ 2 } }{ R } =2\times 10^{ -7 }\times \frac { 5\times 10 }{ 20\times { 10 }^{ -2 } } $$
$$=5\times { 10 }^{ -5 }N/M$$
As both currents are in same direction, force is attractive.
Question 9
1 / -0

The magnitude of the force between a pair of conductors, each of length $$110\ cm$$, carrying a current of $$10\ A$$ each and separated by a distance of $$10 cm$$ is ($$\mu_{0}=4\pi\times 10^{-7}H/m$$)

A
$$55\times 10^{-5}N$$

B
$$44\times 10^{-5}N$$

C
$$33\times 10^{-5}N$$

D
$$22\times 10^{-5}N$$

Solution

$$F=\dfrac {\mu_0 i_1i_2}{2\pi r}(l)$$
$$=\dfrac{2\times 10^{-7}\times 10\times 10\times 1.1}{\cdot 1}$$
$$=22\times 10^{-5} N$$
Question 10
1 / -0

Two conductors each of length $$12m$$ lie parallel to each other in air. The centre to centre distance between the two conductors is $$15\times 10^{-2}m$$ and the current in each conductor is $$300A$$. The force in newton tending to pull the conductors together is:

A
$$14.4 N$$

B
$$1.44 N$$

C
$$144 N$$

D
$$0.144 N$$

Solution

$$F=\dfrac {\mu_0 i_1i_2l}{2\pi r}$$
$$=\dfrac {2\times 10^{-7}\times 300\times 300\times12}{15\times 10^{-2}}$$
$$=1\cdot 44 N$$

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Moving Charges and Magnetism Test - 35

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Moving Charges and Magnetism Test - 35

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Question 1

$$F$$

$$5F$$

$$0.2F$$

$$\sqrt{5}F$$

Solution

Question 2

$$Zero$$

$$\dfrac{mv}{qB}$$

$$\dfrac{2mv}{qB}$$

$$Infinity$$

Solution

Question 3

doubled

remains same

quadrupled

halved

Solution

Question 4

4.85A

zero

$$4.85\times 10^{-2}A$$

$$85\times 10^{-4}A$$

Solution

Question 5

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

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

$$4\times 10^{7}Nm^{-1}$$

$$4\times 10^{-7}Nm^{-1}$$

Solution

Question 6

$$0.2N$$

$$0.4N$$

$$0.6N$$

$$0.8N$$

Solution

Question 7

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

$$3\times 10^{-5}m$$

$$2\times 10^{-5}m$$

$$1\times 10^{-5}m$$

Solution

Question 8

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

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

$$2\times 10^{-5}N/m$$ attractive

$$2\times 10^{-5}N/m$$ repulsive

Solution

Question 9

$$55\times 10^{-5}N$$

$$44\times 10^{-5}N$$

$$33\times 10^{-5}N$$

$$22\times 10^{-5}N$$

Solution

Question 10

$$14.4 N$$

$$1.44 N$$

$$144 N$$

$$0.144 N$$

Solution

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