Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 39

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A magnetic field $$4\times 10^{-3}\hat{k}T$$ exerts a force $$(4\hat{i}+3\hat{j})\times 10^{-10}N$$ on a particle having a charge $$10^{-9}$$C and going in the X-Y plane. The velocity of the particle is

A
$$-75\hat{i}+100\hat{j}$$

B
$$100\hat{i}+75\hat{j}$$

C
$$75\hat{i}+100\hat{j}$$

D
$$100\hat{i}-75\hat{j}$$

Solution

$$\bar{F}=q\left ( \underset{V}{\rightarrow}\times \underset{B}{\rightarrow} \right )$$
$$(4\hat{i}+3\hat{j})\times 10^{-10}=10^{-19}((u\hat{i}+v\hat{j}+w\hat{k})\times (4\times 10^{-3}\hat{k}))$$
$$(4\hat{i}+3\hat{j})\times 25=-u\hat{j}+v\hat{i}$$
comparing the cofficient
$$u=-75; \
v=100$$
Question 2
1 / -0

A charged particle is projected in a magnetic field $$\vec{B}=(3\vec{i}+4\vec{j})\times 10^{-2}T$$ and the acceleration is found to be $$\vec{a}=(x\vec{i}+2\vec{j})m/s^{2}$$. The value of x is :

A
-8/3 $$m/s^{2}$$

B
8/3 $$m/s^{2}$$

C
-4/3 $$m/s^{2}$$

D
4/3 $$m/s^{2}$$

Solution

$$\overline{F}=q(\overline V\times\overline B)$$

So $$\overline F\bot\overline B$$

$$\therefore \overline F\cdot \overline B=0$$

$$\overline ma\cdot\overline B=0$$

$$\overline a\cdot\overline B=0$$

$$(x\overline i+2\overline j)\cdot(3\overline i+4\overline j)=0$$

$$3x+8=0$$

$$x=\dfrac{-8}{3}$$
Question 3
1 / -0

Assertion: If an electron is not deflected while passing through a certain region of space, then only possibility is that there is no magnetic field in this region

Reason: Magnetic force is directly proportional to magnetic field applied

A
Both assertion and reason are true and reason is correct explanation of the assertion.

B
Both assertion and reason are true and reason is not a correct explanation of the assertion.

C
Assertion is true but reason is false

D
Assertion is false but reason is true

Solution

$$\overline{F}=q(\overline{v}\times\overline{B})$$
$$F=qv B \sin\theta$$
if $$\theta=0$$, $$F=0$$
then particle will also not deflect.
And $$F\ \alpha \ B$$
Question 4
1 / -0

A charge having q/m equal to $$10^8$$c/kg and with velocity 3 x$${ 10 }^{ 5 }$$ m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30 with direction of field. Then radius of curvature will be :

A
0.01 cm

B
0.5 cm

C
1 cm

D
2 cm

Solution

$$r=\dfrac { m{ V }_{ \bot } }{ qB } $$
$$r=\left( \dfrac { m }{ q } \right) \left( \dfrac { 3\times { 10 }^{ 5 }\times sin3{ 30 }^{\circ} }{ 0.3 } \right) $$
$$r=\dfrac { 3\times { 10 }^{ 5 } }{ { 10 }^{ 8 }\times 0.3\times 2 } =0.5\times 1{ 0 }^{ -2 }m=0.5cm$$
Question 5
1 / -0

Which of the following statement is not correct about two parallel conductors carrying equal currents in the same direction?

A
each of the conductors will experience a force

B
the two conductors will repel each other

C
there are concentric lines of force around each conductor

D
each of the conductors will move if not prevented from doing so

Solution

When two conductors carrying equal current both the conductor experiences a force which is perpendicular to the wire and are equal. Since both these forces are equal and towards each other, both the current carrying wire will be repelling each other.
Question 6
1 / -0

How will two parallel beams of electron behave while moving in the same direction?

A
repel each other

B
attract each other

C
not interact with each other

D
annihilate each other

Solution

Two parallel beams of electron moving in the same direction can be think of like two current carrying wire in same direction. The situation is shown in the figure. The direction indicated shows that wire 2 will be attracted towards wire 1. In a similar manner, one can show that wire 1 will experience a force due to the magnetic field of wire 2, and that this force will have a magnitude equal to that of F2 but opposite in direction. Thus, wire 1 will be attracted towards wire 2.
Question 7
1 / -0

A tesla is equivalent to a

A
Newton per coulomb

B
Newton per ampere-meter

C
Ampere per newton

D
Newton per ampere-second

Solution

Tesla is the SI unit of magnetic field strength. A particle carrying a charge of coulomb and passing through a magnetic field of 1 tesla at a speed of 1 meter per second perpendicular to said field experiences a force with magnitude 1 newton.
$$T=V.S/m^2=N/A.m$$
Question 8
1 / -0

A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is in which direction?

A
towards south

B
towards east

C
downward

D
upward

Solution
Question 9
1 / -0

A uniform magnetic field exists in the plane of paper pointing from left to right as shown in Figure. In the field an electron and a proton move as shown. In what direction the electron and the proton experience force ?

A
forces both pointing into the plane of paper

B
forces both pointing out of the plane of paper

C
forces pointing into the plane of paper and out of the plane of paper, respectively

D
force pointing opposite and along the direction of the uniform magnetic field respectively

Solution

Force experience by a charged particle having charge q moving with a velocity V in a magnetic field B is given by:
$$\vec F=q({\vec V}\times{\vec B})$$. Now the $$q_ {electron}=-q_ {proton}$$ and $$V_ {electron}=-V_ {proton}$$. The force experience by both of them will be in same direction.
$$V_ {proton}=V_\circ{} \hat{j}; B=B_\circ{} \hat{i}\\ \Rightarrow F=eV_\circ{} B_\circ{}(\hat{-j}\times\hat{i})=-eV_\circ{} B_\circ{}\hat{k}$$
So the force experience by both of them will be pointing into the plane of paper.
Question 10
1 / -0

An electron and a proton move in a uniform magnetic field with same speed $$20 \ m/s$$ perpendicular to the magnetic field ($$1 T$$). What are the forces they will experience ?

A
$$0 N, 0 N$$

B
$$32 \times 10^{-19} N, 32 \times 10^{-19} N$$

C
$$-32 \times 10^{-19} N, 32 \times 10^{-19} N$$

D
$$-32 \times 10^{-18} N, 32 \times 10^{-18} N$$

Solution

The force acting on the charged particle placed in magnetic field is given by,
$$ F = qVB \sin \theta = \pm 1.6 \times 10^{-19} \times 20 \times 1 = \pm 32 \times 10^{-19 } N $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 39

Result

Moving Charges and Magnetism Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$-75\hat{i}+100\hat{j}$$

$$100\hat{i}+75\hat{j}$$

$$75\hat{i}+100\hat{j}$$

$$100\hat{i}-75\hat{j}$$

Solution

Question 2

-8/3 $$m/s^{2}$$

8/3 $$m/s^{2}$$

-4/3 $$m/s^{2}$$

4/3 $$m/s^{2}$$

Solution

Question 3

Both assertion and reason are true and reason is correct explanation of the assertion.

Both assertion and reason are true and reason is not a correct explanation of the assertion.

Assertion is true but reason is false

Assertion is false but reason is true

Solution

Question 4

0.01 cm

0.5 cm

1 cm

2 cm

Solution

Question 5

each of the conductors will experience a force

the two conductors will repel each other

there are concentric lines of force around each conductor

each of the conductors will move if not prevented from doing so

Solution

Question 6

repel each other

attract each other

not interact with each other

annihilate each other

Solution

Question 7

Newton per coulomb

Newton per ampere-meter

Ampere per newton

Newton per ampere-second

Solution

Question 8

towards south

towards east

downward

upward

Solution

Question 9

forces both pointing into the plane of paper

forces both pointing out of the plane of paper

forces pointing into the plane of paper and out of the plane of paper, respectively

force pointing opposite and along the direction of the uniform magnetic field respectively

Solution

Question 10

$$0 N, 0 N$$

$$32 \times 10^{-19} N, 32 \times 10^{-19} N$$

$$-32 \times 10^{-19} N, 32 \times 10^{-19} N$$

$$-32 \times 10^{-18} N, 32 \times 10^{-18} N$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.