Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

A charged particle moving in a magnetic field experiences, some magnetic force because

A
Moving charged particle produce electric field

B
Moving charged particle produce magnetic field

C
Magnetic field do not allow the charged particle to pass through it.

D
none of these

Solution
Question 2
1 / -0

Motion of charges is noting but :

A
Electric current

B
magnetic effect

C
heating effect

D
all of the above

Solution

Due to motion of charges through a conductor, the current of charges passing through its cross-section changes. It causes electric current $$\left( ie. \dfrac{da}{dt}\right)$$. Because of this current, heating effect and magnetic effect are produced in accordance with Joules and Bio Savaert's law resp.
Option $$-D$$ is correct.
Question 3
1 / -0

If a proton enters perpendicularly a magnetic field with velocity v, then time period of revolution is T. If proton enters with velocity $$2$$v, then time period will be?

A
T

B
$$2$$T

C
$$3$$T

D
$$ \frac{T}{\sqrt{2}} $$

Solution
Question 4
1 / -0

A conductor of length $$1\ m$$ and carrying current of $$1\ A$$ is placed at an angle $$45 ^ { \circ }$$ to the magnetic field of 1 oersted. The force acting on conductor is

A
$$\dfrac { 10 ^ { - 4 } } { \sqrt { 2 } } N$$

B
$$\dfrac { 10 ^ { - 4 } } { \sqrt { 3 } } N$$

C
$$\dfrac { 10 ^ { - 2 } } { \sqrt { 3 } } N$$

D
$$\dfrac { 10 ^ { - 2 } } { \sqrt { 2 } } N$$

Solution

It is given that,

Length, $$l=1\,m$$

Current, $$I=1\,A$$

Magnetic field, $$B=1\,oersted={{10}^{-4}}\,Kg/A{{s}^{2}}$$

Magnetic force,

$$ F=I\left( L\times B \right) $$

$$ F=1\left( 1\times {{10}^{-4}}\sin 45 \right) $$

$$ F=\dfrac{{{10}^{-4}}}{\sqrt{2}} $$
Question 5
1 / -0

A charged particle enters in a magnetic field $$H$$ with its initial velocity making an angle of $$45^o$$ with $$H$$. The path of the particle will be

A
an ellipse

B
straight line

C
a circle

D
a helical

Solution

The component of velocity perpendicular to $$H$$ will make the motion circular while that parallel to $$H$$ will make it move along a straight line. The two together will make the motion helical.
Question 6
1 / -0

The magnitude of net magnetic force on wire frame as shown below is ( the frame is kept in a perpendicular magnetic field $$ \bar B $$)

A
$$ BI_0L $$

B
$$ \frac {BI_0L }{2} $$

C
$$ \frac {3}{2} BI_0L $$

D
$$ 2 BI_0L $$

Solution

Given:
The wire system with there length

Solution:
The force on the longer sides will cancel out each other and the force on the smaller side will be =BI$$\circ$$L

So,correct opt: A
Question 7
1 / -0

A proton and an alpha particle enter the same magnetic field which is perpendicular to their velocity. If they have same kinetic energy then ratio of radii of their circular path is

A
$$1:1$$

B
$$1:2$$

C
$$2:1$$

D
$$1:4$$

Solution

A proton and an alpha particle enter the same magnetic field which is perpendicular to their velocity and have the same kinetic energy.

Given,

$$m_{\alpha}=4m_p$$

$$q_{\alpha}=2q_p$$

The radius of the circular path is given by

$$r=\dfrac{\sqrt{2mE}}{qB}$$. . . . . . . .(1)

$$r\propto \dfrac{\sqrt{m}}{q}$$

$$\dfrac{r_{p}}{r_{\alpha}}=\sqrt{\dfrac{m_p}{m_{\alpha}}}.\dfrac{q_{\alpha}}{q_p}$$

$$\dfrac{r_p}{r_{\alpha}}=\sqrt{\dfrac{m_p}{4m_p}}.\dfrac{2q_p}{q_p}$$

$$\dfrac{r_p}{r_{\alpha}}=2\sqrt{\dfrac{1}{4}}$$

$$\dfrac{r_p}{r_{\alpha}}=1$$

$$r_p:r_{\alpha}=1:1$$

The correct option is A.
Question 8
1 / -0

An electron moving along a horizontal line with respect to the observer and towards the north. Due to each magnetic field, it deflects towards the east then position observer:-

A
In NHS

B
In SHS

C
At Equator

D
None

Solution

The magnetic field lines from south to north,
$$\vec F_m=q(\vec v\times \vec B)$$
The direction of electron velocity is in the negative y-axis.
$$\hat i=-\hat j\times -\hat k$$
The magnetic field along the direction of $$-\hat k$$
It is possible only in NHS.
The correct option is A.
Question 9
1 / -0

There are a number of hydrogen like atoms is free space. Initially all the atoms are stationary and in a particular excited level $$A$$. They undergo transitions from the initial level $$A$$ to the ground state. The recoiling atom enters a region of constant magnetic field. If the radii of the paths of the recoiling atoms is observed to have only three different values, then

A
Magnetic moment of the atoms in the initial state $$A$$ is $$\dfrac {5eh}{4\pi m}$$

B
The radii must be in the ratio $$5 : 27 : 32$$

C
The angular momenta of the electrons in the initial state of the emitting atoms is $$\dfrac {3h}{2\pi}$$

D
The principal quantum number of the level $$A$$is given by $$n = 3$$
Question 10
1 / -0

A stationary magnet does not interact with

A
iron rod

B
moving charge

C
moving magnet

D
stationary charge

Solution

Magnetic force on a stationary charge is always zero. In order to experience magnetic force the charge must have non zero velocity.

Magnetic Force, $$F_B = q (\vec{v} \times \vec{B})$$

If $$\vec{v} = 0 $$ then $$F_B = 0 $$

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Moving Charges and Magnetism Test - 62

Result

Moving Charges and Magnetism Test - 62

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Question 1

Moving charged particle produce electric field

Moving charged particle produce magnetic field

Magnetic field do not allow the charged particle to pass through it.

none of these

Solution

Question 2

Electric current

magnetic effect

heating effect

all of the above

Solution

Question 3

T

$$2$$T

$$3$$T

$$ \frac{T}{\sqrt{2}} $$

Solution

Question 4

$$\dfrac { 10 ^ { - 4 } } { \sqrt { 2 } } N$$

$$\dfrac { 10 ^ { - 4 } } { \sqrt { 3 } } N$$

$$\dfrac { 10 ^ { - 2 } } { \sqrt { 3 } } N$$

$$\dfrac { 10 ^ { - 2 } } { \sqrt { 2 } } N$$

Solution

Question 5

an ellipse

straight line

a circle

a helical

Solution

Question 6

$$ BI_0L $$

$$ \frac {BI_0L }{2} $$

$$ \frac {3}{2} BI_0L $$

$$ 2 BI_0L $$

Solution

Question 7

$$1:1$$

$$1:2$$

$$2:1$$

$$1:4$$

Solution

Question 8

In NHS

In SHS

At Equator

None

Solution

Question 9

Magnetic moment of the atoms in the initial state $$A$$ is $$\dfrac {5eh}{4\pi m}$$

The radii must be in the ratio $$5 : 27 : 32$$

The angular momenta of the electrons in the initial state of the emitting atoms is $$\dfrac {3h}{2\pi}$$

The principal quantum number of the level $$A$$is given by $$n = 3$$

Question 10

iron rod

moving charge

moving magnet

stationary charge

Solution

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