Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The force felt by an electron on entering into a magnetic field is independent of its

A
Charge

B
Strength of the field

C
Mass

D
Direction of its velocity

Solution

$$ \vec{F_B} = q \vec{v} \times \vec{B}$$
independent of mass
Question 2
1 / -0

You are sitting in a room in which uniform magnetic field is present in vertically downward direction. When an electron is projected in horizontal direction, it will be moving in circular path with constant speed

A
clockwise in vertical plane

B
clockwise in horizontal plane

C
anticlockwise in vertical plane

D
anticlockwise in horizontal plane

Solution

by $$F = q\vec{v} \times \vec{B}$$
where $$q$$ is $$-ve$$
therefore, it will be moving in a circular path with constant speed clockwise in horizontal plane
Question 3
1 / -0

A neutron , a proton, an electron and an $$\alpha $$ particle enter perpendicular uniform magnetic field, with the same uniform velocity. The path of electron in the following figure will be :

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

The neutron goes undeflected, the $$\alpha $$ and proton being both positively charged moves towards left and electron being negatively charged moves towards right. or we can also use right hand to find the direction of force using formula $$ F = q(V\times B)$$ but it moves right as $$q$$ is $$-ve.$$
Question 4
1 / -0

The magnetic field $$\overline{dB}$$ due to a small current element dl at a distance $$\vec{r}$$ and carrying current ‘i’ is

A
$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$$

B
$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{2}} \right )$$

C
$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$$

D
$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{3}} \right )$$

Solution

Bio-savert law - apply directly the statement of law
Question 5
1 / -0

For a given distance from a current element, the magnetic induction is maximum at an angle measured with respect to axis of the current. The angle is :

A
$$\dfrac{3\pi}{ 4}$$

B
$$\dfrac{\pi }{4}$$

C
$$\dfrac{\pi} {2}$$

D
$$2\pi $$

Solution

Magnetic field due to small current element is given as
$$B=\dfrac{\mu_o}{4\pi}\dfrac{\vec{idl}\times \hat{r}} {r^2}$$
$$B=\dfrac{\mu_o}{4\pi}\dfrac{idlr \sin\theta}{r^3}$$
$$\sin\theta=1$$ for $$\theta=\dfrac{\pi}{2}$$
Question 6
1 / -0

A positive charged particle enters at the middle as shown in the figure with $$10^{5}\ m/s$$ will bend:

A
towards 1 A wire.

B
upwards the plane of wires.

C
towards 3 A wire.

D
down wards the plane of wires.

Solution

According to right hand thumb rule, magnetic field due to left wire is inside the paper while the magnetic field due to the right wire is outside the paper. As the point is in middle and current is more in left wire, net magnetic field is inside the wire. Using the Fleming's left hand rule, it can be deduced that the force on the particle is in the left direction.
Question 7
1 / -0

A magnetic field exerts no force on

A
a stream of electrons

B
a stream on protons

C
unmagnetised piece of iron

D
stationary charge.

Solution

$$F=q(\overrightarrow{v}\times \overrightarrow{B})$$
if $$v=0\rightarrow F=0$$
Question 8
1 / -0

Magnetic field at a point on the line of current carrying conductor is

A
maximum

B
infinity

C
zero

D
finite value

Solution

because angle between line and distance becomes zero.
$$\therefore\overrightarrow{l}\times\widehat{r}$$ becomes zero.
Question 9
1 / -0

If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field, the orbit will be :

A
a straight line

B
an ellipse

C
a circle

D
a helix

Solution

If it enters pependicular to the magnetic field, the force applied on it will be perpendicular to it's actual velocity.
$$\therefore$$ it will move on circular path, [Right hand thumb rule]
Question 10
1 / -0

A cathode ray beam is bent into an arc of a circle of radius $$0.02 m$$ by a field of magnetic induction $$4.55 mT$$. The velocity of electrons is:
(Given $$e=1.6\times 10^{-19}c$$ and $$m=9.1\times 10^{-31}kg$$)

A
$$2\times 10^{7}m/s$$

B
$$3\times 10^{7}m/s$$

C
$$1.6\times 10^{7}m/s$$

D
$$3.2\times 10^{7}m/s$$

Solution

Radius in Magnetic field $$= \dfrac{mV}{eB}$$

$$\Rightarrow V = \dfrac{Ber}{m}$$

$$\Rightarrow V = \dfrac{4.55 \times 10^{-3} \times 1.6 \times 10^{-19} \times 0.02}{9.1 \times 10^{-31}}$$

$$\Rightarrow V = 1.6 \times 10^{7}m/s$$

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Moving Charges and Magnetism Test - 32

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Moving Charges and Magnetism Test - 32

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Question 1

Charge

Strength of the field

Mass

Direction of its velocity

Solution

Question 2

clockwise in vertical plane

clockwise in horizontal plane

anticlockwise in vertical plane

anticlockwise in horizontal plane

Solution

Question 3

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 4

$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$$

$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{2}} \right )$$

$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$$

$$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{3}} \right )$$

Solution

Question 5

$$\dfrac{3\pi}{ 4}$$

$$\dfrac{\pi }{4}$$

$$\dfrac{\pi} {2}$$

$$2\pi $$

Solution

Question 6

towards 1 A wire.

upwards the plane of wires.

towards 3 A wire.

down wards the plane of wires.

Solution

Question 7

a stream of electrons

a stream on protons

unmagnetised piece of iron

stationary charge.

Solution

Question 8

maximum

infinity

zero

finite value

Solution

Question 9

a straight line

an ellipse

a circle

a helix

Solution

Question 10

$$2\times 10^{7}m/s$$

$$3\times 10^{7}m/s$$

$$1.6\times 10^{7}m/s$$

$$3.2\times 10^{7}m/s$$

Solution

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