Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 34

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

An electron is projected in the same direction of uniform magnetic field. Then

A
the electron turns to right

B
the electron turns to left

C
the electron velocity remains constant

D
the electron velocity decreases in magnitude

Solution

$$\overrightarrow{V}||\overrightarrow{B}, Sin\theta=0$$
no force by field acts on it.
Question 2
1 / -0

A charged particle moving in a magnetic field experiences a resultant force

A
in the direction opposite to that of the field.

B
in the direction opposite to that of its velocity

C
in the direction perpendicular to both field &

its velocity

D
in the direction parallel to the field

Solution

$$\overrightarrow{F}=q(\overrightarrow{V}\times\overrightarrow{B})$$
F is perpendicular to both velocity and field.
Question 3
1 / -0

A proton and an electron enter a region with equal speed in which a magnetic field is suddenly switched on. The force experienced by them are

A
equal and opposite

B
different in magnitude but same in direction

C
in the ratio of 1840

D
same in magnitude and direction

Solution

charge is equal in magnitude but opposite in sign
$$\therefore$$ forces are equal in magnitude but opposite in direction
Question 4
1 / -0

When a charged particle moves in a uniform magnetic field

A
it gains energy from the field

B
it loses energy to the field

C
it neither gains nor changes energy and momentum

D
momentum changes but not energy

Solution

Charged particle moving in a magnetic field conserve it's energy but due to the change in the direction of the particle, it's momentum changes.
Question 5
1 / -0

A proton enters a magnetic field with a velocity of $$2.5\times 10^{7}ms^{-1}$$ making an angle $$30^{0}$$ with the magnetic field. The force on the proton is $$(B=25T)$$

A
$$1.25\times 10^{-11}N$$

B
$$2.5\times 10^{-11}N$$

C
$$5.0\times 10^{-11}N$$

D
$$7.5\times 10^{-11}N$$

Solution

$$\bar F=q(\vec {v}\times \vec {B})$$
$$F-qvB sin\theta$$
$$F=1\cdot 6\times 10^{-19}\times 2\cdot 5\times 10^7\times 25\times sin 30$$
$$F=5\times 10^{-11} N$$
Question 6
1 / -0

A particle of mass $$M$$ and charge $$Q$$ moving with velocity $$\vec{v}$$ describes a circular path of radius $$R$$ when subjected to a uniform transverse magnetic field of induction $$\vec{B}$$. The work done by the field when the particle completes one full circle is

A
$$MV^{2}R^2 $$

B
$$0$$

C
$$VBQR$$

D
$$2\pi VBQR$$

Solution

$$\bar F=q(\bar {V}\times \bar{B})$$
Here $$\bar F$$ is perpendicular to the velocity of particle.
So $$P=\bar {F}\cdot \bar {V}$$
$$=0$$
So the workdone by the particle $$=0$$
Question 7
1 / -0

A proton is rotating along a circular path with kinetic energy K in a uniform magnetic field B.If the magnetic field is made four times, the kinetic energy of rotation of proton is

A
16K

B
8K

C
4K

D
K

Solution

The force acting on particle moving with speed $$v$$ is given by
$$F=qvB$$
This provides the centripetal acceleration to the particle.
hence $$qvB=\dfrac{mv^2}{r}$$
$$\implies v=\dfrac{qBr}{m}$$
Hence the kinetic energy=$$K=\dfrac{1}{2}mv^2\propto B^2$$
Hence when magnetic field becomes 4 times, the kinetic energy of proton becomes 16 times.
Question 8
1 / -0

A charged particle is moving with velocity $$v$$ in a magnetic field of induction $$B$$. The force on the particle will be maximum when

A
$$v$$ and $$B$$ are in the same direction

B
$$v$$ and $$B$$ are in Opposite direction

C
$$v$$ and $$B$$ are perpendicular

D
$$v$$ and $$B$$ are at an angle of $$45^{o}$$

Solution

Force $$= q(\vec{V}\times \vec{B})$$
$$= q VB \sin 0$$
Where, $$0$$ is angle between velocity and magnetic field.
So, $$\sin 0$$ is maximum when $$0$$ is $$90^o$$ or velocity is perpendicular to magnetic field.
Question 9
1 / -0

A particle of mass $$0.6$$ $$g$$ and having a charge of $$25$$ $$nC$$ is moving horizontally with a uniform velocity $$1.2 \times10^{4}$$ $$m/s$$ in a uniform magnetic field, then the value of the magnetic induction is (Take $$g=10 \ m/s^2$$)

A
$$zero$$

B
$$10 T$$

C
$$20 T$$

D
$$200 T$$

Solution

Particle is moving with uniform velocity therefore no force is acting on the particle.
So $$mg- qvB=0$$
$$0\cdot 6\times 10^{-3}\times 10=25\times 10^{-9}\times 1\cdot 2\times 10^4\times B$$
$$B=20 T$$
Question 10
1 / -0

A particle consisting of two electrons is moving in a magnetic field of $$(3i+2j) T$$ with a velocity $$5 \times 10^{5}\hat{i}m/s$$. The magnetic force acting on the particle will be

A
$$3.2\times 10^{-13}\hat{k} \ dyne$$

B
$$3.2\times 10^{13}\hat{k} \ N$$

C
$$zero$$

D
$$3.2\times 10^{-13}\hat{k} \ N$$

Solution

$$\vec F=q(\vec v\times \vec B)$$
$$\bar F=3\cdot 2\times 10^{-19}(5\times 10^5 \hat i\times (3 \hat i+2 \hat j))$$
$$=3\ . 2\times 10^{-13} \ \hat k \ N$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 34

Result

Moving Charges and Magnetism Test - 34

Score

-

Rank

-

SHARING IS CARING

Question 1

the electron turns to right

the electron turns to left

the electron velocity remains constant

the electron velocity decreases in magnitude

Solution

Question 2

in the direction opposite to that of the field.

in the direction opposite to that of its velocity

in the direction perpendicular to both field &its velocity

in the direction parallel to the field

Solution

Question 3

equal and opposite

different in magnitude but same in direction

in the ratio of 1840

same in magnitude and direction

Solution

Question 4

it gains energy from the field

it loses energy to the field

it neither gains nor changes energy and momentum

momentum changes but not energy

Solution

Question 5

$$1.25\times 10^{-11}N$$

$$2.5\times 10^{-11}N$$

$$5.0\times 10^{-11}N$$

$$7.5\times 10^{-11}N$$

Solution

Question 6

$$MV^{2}R^2 $$

$$0$$

$$VBQR$$

$$2\pi VBQR$$

Solution

Question 7

16K

8K

4K

K

Solution

Question 8

$$v$$ and $$B$$ are in the same direction

$$v$$ and $$B$$ are in Opposite direction

$$v$$ and $$B$$ are perpendicular

$$v$$ and $$B$$ are at an angle of $$45^{o}$$

Solution

Question 9

$$zero$$

$$10 T$$

$$20 T$$

$$200 T$$

Solution

Question 10

$$3.2\times 10^{-13}\hat{k} \ dyne$$

$$3.2\times 10^{13}\hat{k} \ N$$

$$zero$$

$$3.2\times 10^{-13}\hat{k} \ N$$

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.

in the direction perpendicular to both field &

its velocity