Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 58

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

Ampere's circuital law is given by:

A
$$\oint \bar{H} . \bar{dl} = \mu _0 I_{enc}$$

B
$$\oint \bar{B} . \bar{dl} = \mu _0 I_{enc}$$

C
$$\oint \bar{B} . \bar{dl} = \mu _0 I$$

D
$$\oint \bar{H} . \bar{dl} = \mu _0 I$$

Solution

The line integral of the magnetic field of induction $$\vec B$$ around any closed path in free space is equal to absolute permeability of free space $$μ_o$$ times the total current flowing through area bounded by the path.
Ampere's circuital law is given by:
$$\oint \vec B.d\vec l=\mu_0 I_{enc}$$
The correct option is B.
Question 2
1 / -0

Which of the following material is used in making the core of a moving coil galvanometer?

A
Copper

B
Nickel

C
Iron

D
Both (a) and (b).

Solution

Soft iron is used in making the core of moving coil galvanometer, because it has high initial permeability and low hysteresis loss.
Question 3
1 / -0

At a place an electric field and a magnetic field are in downward direction. There an electron moves in downward direction. Hence this electron.

A
Will bend towards left

B
Will bend towards right

C
Will gain velocity

D
Will lose velocity

Solution

When a charge enters a region where bothelectric and magnetic field exists simultaneously, Lorentz force acts on it and is given by $$\vec{F} = q[\vec{E} + (\vec{v} \times \vec{B} )]$$
Since $$\vec v$$ and $$\vec B$$ are in same direction, $$\vec V \times \vec B = vB\sin0^0 = 0$$ ,So, only electric force will act on it in upward direction $$(\vec F = -q\vec E)$$ which is in opposite direction of the velocity.Hence this electron will lose velocity.
Therefore, D is correct option.
Question 4
1 / -0

A conducting square loop of side $$L$$ and resistance $$R$$ moves in its plane with a uniform velocity $$v$$ perpendicular to one of its sides. A magnetic induction $$B$$, constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere, then the current induced in the loop is

A
$$BLv/R$$ clockwise

B
$$BLv/R$$ anticlockwise

C
$$2 BLv/R$$ anticlockwise

D
$$0$$

Solution

Polarity induced $$X$$ is shown in fig.
Due to same voltage on same side no current flows.
Question 5
1 / -0

A charged particle moves through a magnetic field perpendicular to its direction. Then

A
the momentum changes but the kinetic energy is constant

B
both momentum and kinetic of the particle are not constant

C
both, momentum and kinetic energy of the particle are constant

D
kinetic energy changes but the momentum is constant

Solution
Question 6
1 / -0

Amperes circuital law is given by

A
$$\oint\bar H. \bar {dl} =\mu_0 I_{enc}$$

B
$$\oint\bar B. \bar {dl} =\mu_0 I$$

C
$$\oint\bar B. \bar {dl}=\mu_0 J$$

D
$$\oint\bar H. \bar {dl}=\mu_0 J$$

Solution

Ampere 's circuital law states that relationship between thecurrent and the magnetic field
$$\phi\overrightarrow { B } \overrightarrow { dl } =\mu_0 I$$
B is correct option
Question 7
1 / -0

An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is (q = 1.6 X $$10^{-19}$$ C, $$m_e$$ = 9.1 X $$10^{-31}$$ kg)

A
2.58 X $$10^{-4}$$m

B
3.58 X $$10^{-4}$$m

C
2.58 X $$10^{-3}$$m

D
3.58 X $$10^{-3}$$m

Solution

$$\displaystyle E = \frac{1}{2} mv^2 \Rightarrow v = \sqrt{ \dfrac{2E}{m} }$$
$$\displaystyle r = \frac{mv}{Be} = \frac{m}{Be} \sqrt{ \dfrac{2E}{m} } = \dfrac{\sqrt{2mE}}{Be}$$
$$\displaystyle r = \frac{\sqrt{2 \times 1800 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31}}}{1.6 \times 10^{-19} \times 0.4} = 3.58 \times 10^{-4} m$$
Question 8
1 / -0

If an electron is moving with velocity $$\bar{v}$$ produces a magnetic field $$\bar{B}$$, then

A
the direction of field $$\bar{B}$$ will be same as the direction of velocity $$\bar{v}$$

B
the direction of field $$\bar{B}$$ will be opposite as the direction of velocity $$\bar{v}$$

C
the direction of field $$\bar{B}$$ will be perpendicular as the direction of velocity $$\bar{v}$$

D
the direction of field $$\bar{B}$$ does not depend upon the direction of velocity $$\bar{v}$$

Solution

According to Biot-Savart's law, the magnetic field
$$\displaystyle \overset{\rightarrow}{B} = \frac{\mu_o}{4 \pi} . \frac{q (\overset{\rightarrow}{v} \times \overset{\rightarrow}{r} ) }{r^3}$$
The direction of $$\overset{\rightarrow}{B}$$ will be along $$\overset{\rightarrow}{v} \times \overset{\rightarrow}{r}$$ i.e. perpendicular to the plane containing $$\overset{\rightarrow}{v}$$ and $$\overset{\rightarrow}{r}$$.
Question 9
1 / -0

Biot-Savart law indicates that the moving electrons (velocity $$\bar v$$ ) produce a magnetic field $$\bar B$$ such that:

A
$$\bar B \perp \bar v$$

B
$$\bar B \parallel \bar v$$

C
it obeys inverse cube law.

D
it is along the line joining the electron and point of observation.

Solution

Magnetic field produced by charges moving with velocity $$\bar v$$, at a distance r is $$ \bar B$$ = $$\left ( \dfrac{\mu_0}{4\pi } \right )$$.q$$\dfrac{\bar v \times \hat r}{r^2}$$
Therefore $$\bar B \perp \bar v$$
Question 10
1 / -0

Fill in the blanks
A charge is moving through a magnetic field. The force acting on the charge is maximum when the angle between the velocity of charge and the magnetic field is ______

A
$$\pi$$

B
$$\pi/3$$

C
$$\pi/2$$

D
$$\pi/4$$

Solution

We know,

$$\overrightarrow{F_B}=q\overrightarrow{v}\times\overrightarrow B=qvB\sin\theta$$
Where: $$\theta$$ = angle between v and B ,

Now for $$\overrightarrow F$$ to be maximum, $$\sin\theta$$ should be max,

i.e $$\theta =\dfrac{\pi}{2}$$

Option $$\textbf C$$ is the correct answer.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 58

Result

Moving Charges and Magnetism Test - 58

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\oint \bar{H} . \bar{dl} = \mu _0 I_{enc}$$

$$\oint \bar{B} . \bar{dl} = \mu _0 I_{enc}$$

$$\oint \bar{B} . \bar{dl} = \mu _0 I$$

$$\oint \bar{H} . \bar{dl} = \mu _0 I$$

Solution

Question 2

Copper

Nickel

Iron

Both (a) and (b).

Solution

Question 3

Will bend towards left

Will bend towards right

Will gain velocity

Will lose velocity

Solution

Question 4

$$BLv/R$$ clockwise

$$BLv/R$$ anticlockwise

$$2 BLv/R$$ anticlockwise

$$0$$

Solution

Question 5

the momentum changes but the kinetic energy is constant

both momentum and kinetic of the particle are not constant

both, momentum and kinetic energy of the particle are constant

kinetic energy changes but the momentum is constant

Solution

Question 6

$$\oint\bar H. \bar {dl} =\mu_0 I_{enc}$$

$$\oint\bar B. \bar {dl} =\mu_0 I$$

$$\oint\bar B. \bar {dl}=\mu_0 J$$

$$\oint\bar H. \bar {dl}=\mu_0 J$$

Solution

Question 7

2.58 X $$10^{-4}$$m

3.58 X $$10^{-4}$$m

2.58 X $$10^{-3}$$m

3.58 X $$10^{-3}$$m

Solution

Question 8

the direction of field $$\bar{B}$$ will be same as the direction of velocity $$\bar{v}$$

the direction of field $$\bar{B}$$ will be opposite as the direction of velocity $$\bar{v}$$

the direction of field $$\bar{B}$$ will be perpendicular as the direction of velocity $$\bar{v}$$

the direction of field $$\bar{B}$$ does not depend upon the direction of velocity $$\bar{v}$$

Solution

Question 9

$$\bar B \perp \bar v$$

$$\bar B \parallel \bar v$$

it obeys inverse cube law.

it is along the line joining the electron and point of observation.

Solution

Question 10

$$\pi$$

$$\pi/3$$

$$\pi/2$$

$$\pi/4$$

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.