Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The correct figure showing Fleming's Left Hand Rule is :

A

B

C

D

Solution

Whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in the direction perpendicular to both the directions of the directions of the current and the magnetic field.

Question 2

1 / -0

Choose the correct alternative which matches second and third column with first column:

Column I	Column II	Column III
(I) Magnetic field is produced near current carrying conductor	(A) Right hand thumb rule	(a) Micheal Faraday
(II) Electric current is generated in a conductor moving in a magnetic field.	(B) Fleming's right hand rule	(b) Hans Oersted

(I)-(B)-(a),(II)-(B)-(b)

(I)-(A)-(b),(II)-(B)-(b)

(I)-(B)-(b),(II)-(A)-(a)

(I)-(A)-(b),(II)-(B)-(a)

Solution

Magnetic field around conductor is given by Right hand thumb rule and it is associated with oersted while generation of current is given by Fleming's right hand rule and stated by Micheal Faraday.

Question 3
1 / -0

The force acting on a charge q moving with a velocity V in a magnetic field of induction B is given by :

A
$$\dfrac{q}{\vec{V}\times \vec{B}}$$

B
$$\dfrac{\vec{V}\times \vec{B}}{q}$$

C
$$q(\vec{V}\times \vec{B})$$

D
$$(\vec{V}.\vec{B})q$$

Solution

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule . The force is given by the charge times the vector product of velocity and magnetic field.
Question 4
1 / -0

If an electron of velocity is $$2i+4j$$ is subjected to magnetic field of $$4k$$, then, for the electron ,

A
speed will change

B
path will change

C
velocity is Constant

D
momentum is Constant

Solution

$$F = q(\vec{V}\times \vec{B})$$
$$= e[(2i+4j)\times 4k]$$
$$= e[-8j+16i]$$ (By cross-product rule)
So, it has non-zero force, so it velocity direction changes so, path changes, velocity changes, momentum changes but magnitude of velocity is constant, so, speed remains constant.
Question 5
1 / -0

A vertical straight conductor carries a current vertically upwards. A point $$P$$ lies to the east of it at a small distance and another point $$Q$$ lies to the west at the same distance the magnetic field at $$P$$ is :

A
greater than at $$Q$$

B
same as at $$Q$$

C
less than at $$Q$$

D
greater or less than at $$Q$$ depending upon the strength of current

Solution

As per Biot Savart Law, Magnetic field at a point is inversely proportional to square of distance from the current carrying conductor. Therefore magnitude of magnetic field is same at both points P and Q, irrespective of their position from the conductor.
Question 6
1 / -0

If a charged particle is projected perpendicular to a uniform magnetic field, then
a) it revolves in circular path
b) its K.E. remains constant
c) its momentum remains constant
d) its path is spiral

A
only a, b are correct

B
only a, c are correct

C
only b, d are correct

D
only a, d are correct

Solution

If a charged particle is projected perpendicular to a uniform magnetic field, then particle revolves in a circular path and its kinetic energy remains constant since there is no work done by magnetic force. Since a circular path direction of particle changes, so, momentum also changes.
Question 7
1 / -0

A proton moving in a straight line enters a strong magnetic field along the field direction then its path and velocity will be

A
path is circular but speed constant

B
path is same but velocity increases

C
path is same and velocity remains constant

D
path is same and motion is retarded

Solution

$$F = q (\vec{V}\times \vec{B})$$
$$= q VB sin 0$$
$$= q VB sin 0^0$$ $$(\because \text{V and B has same direction})$$
$$= 0$$ $$(\because sin0^0 = 0)$$
So, there is no force, so, path is same and velocity is constant.
Question 8
1 / -0

A proton is traveling due north. If a uniform magnetic field is applied vertically down then the particle
a) deflected towards west
b) revolves in anti clock wise direction
c) acquires acceleration
d) comes to rest

A
a, b, c are correct

B
a, d are wrong

C
b, d are wrong

D
b, c, d are correct

Solution

$$\vec{F} = q(\vec{V}\times \vec{B})$$
So, $$\vec{V}\times \vec{B}$$ direction will be in west by right hand thumb rule. So, it will revolves in anti-clockwise direction and it acquires radial acceleration.
Question 9
1 / -0

A charged particle is projected perpendicular to a uniform magnetic field. Arrange the following in correct order of occurrence :
A) The direction of momentum of the particle changes
B) The particle experiences a force
C) The particle revolves in a circular path

A
A,B,C

B
B,A,C

C
C,B,A

D
C,A,B

Solution

When particle is projected in uniform magnetic field a force = $$q(\vec{V}\times \vec{B})$$ acts which result in change in direction of momentum of the particle and particle starts revolving in a circular path.
So, order is B, A, C.
Question 10
1 / -0

A charge moving with velocity v in X - direction is subjected to a field of magnetic induction in the negative X direction. As a result the charge will

A
retard along X-axis

B
move along a helical path around X axis

C
remain unaffected

D
starts moving in a circular path

Solution

$$F = q (\vec{V}\times \vec{B})$$
$$= q VB sin 0$$
$$= q VB sin(180^0)$$ $$(\because\text { velocity is in X-direction and B is in -X-direction})$$
$$= 0$$ $$(\because sin180^0 = 0)$$
So, charge will be more unaffected.

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Moving Charges and Magnetism Test - 23

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Moving Charges and Magnetism Test - 23

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Question 1

Solution

Question 2

(I)-(B)-(a),(II)-(B)-(b)

(I)-(A)-(b),(II)-(B)-(b)

(I)-(B)-(b),(II)-(A)-(a)

(I)-(A)-(b),(II)-(B)-(a)

Solution

Question 3

$$\dfrac{q}{\vec{V}\times \vec{B}}$$

$$\dfrac{\vec{V}\times \vec{B}}{q}$$

$$q(\vec{V}\times \vec{B})$$

$$(\vec{V}.\vec{B})q$$

Solution

Question 4

speed will change

path will change

velocity is Constant

momentum is Constant

Solution

Question 5

greater than at $$Q$$

same as at $$Q$$

less than at $$Q$$

greater or less than at $$Q$$ depending upon the strength of current

Solution

Question 6

only a, b are correct

only a, c are correct

only b, d are correct

only a, d are correct

Solution

Question 7

path is circular but speed constant

path is same but velocity increases

path is same and velocity remains constant

path is same and motion is retarded

Solution

Question 8

a, b, c are correct

a, d are wrong

b, d are wrong

b, c, d are correct

Solution

Question 9

A,B,C

B,A,C

C,B,A

C,A,B

Solution

Question 10

retard along X-axis

move along a helical path around X axis

remain unaffected

starts moving in a circular path

Solution

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