Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury when a current from the battery is started in the coil through the mercury

A
the wire oscillates

B
the wire continues making contact

C
the wire breaks contact just when the current is passed

D
the mercury will expand by heating due to passage of current

Solution

The windings of the metal helix carry currents in the same direction, therefore they experience attractive forces, pulling the lower end of metal wire out of mercury. When it happens, the circuit breaks, the current disappears and hence the force of attraction disappears. As a result of it, the wire helix regains its original position, completing the circuit again. The process is repeated time and again. Due to it, the oscillatory motion is set up.
So the correct option is (A)
Question 2
1 / -0

The magnetic field due to a current element is independent of :

A
current through it

B
distance from it

C
its length

D
nature of meterial

Solution

$$B=\dfrac{\mu_o}{4\pi}\dfrac{i\overrightarrow{dl}\times\overrightarrow{r}}{r^3}$$

So it depends on all three : current , distance and length
but not on nature of material
Question 3
1 / -0

Imagine that you are sitting in a room with your back to one wall and that an electron beam traveling horizontally from the back wall towards the front one is deflected towards the right. What is the direction of the magnetic induction field that exists in the room?

A
vertically upwards

B
vertically downwards

C
horizontal and perpendicular to the direction of

motion of the electron beam

D
horizontal and parallel to the direction of motion

of the electron beam

Solution

The magnetic field has to be vertically downward for bending electron beam towards right
Question 4
1 / -0

An electron and a proton enter a magnetic field with equal velocities. The particle that experiences more force is

A
electron

B
proton

C
both experience same force

D
it cannot be predicted.

Solution

as charges and velocities are same
$$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$$
will be same, but in opposit direction
Question 5
1 / -0

If the direction of the initial velocity of a charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be:

A
a straight line

B
an ellipse

C
a circle

D
a helix.

Solution

The perpendicular component will be responsible for circular motion and parallel component will take it along the magnetic field. Considering both the phenomena, the resultant motion will be helical.
Question 6
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
remain unaffected

B
start moving in a circular path in $$Y-Z$$ plane

C
retard along $$X-$$axis

D
move along a helical path around $$X -$$ axis

Solution

velocity $$V$$ is in $$X$$ direction and magnetic field $$B$$ is in negative $$X$$ direction hence angle between the velocity and magnetic field is $$180^o$$.
In given case force on moving charge due to magnetic field is given by,
$$F=qVBsin180^o=0$$
Hence charge remains unaffected.
Question 7
1 / -0

An electron of mass m is accelerated through a potential difference of V and then it enters a magnetic field of induction B normal to the lines of force. Then the radius of the circular path is

A
$$\sqrt{\dfrac{2eV}{m}}$$

B
$$\sqrt{\dfrac{2Vm}{eB^{2}}}$$

C
$$\sqrt{\dfrac{2Vm}{eB}}$$

D
$$\sqrt{\dfrac{2Vm}{e^{2}B}}$$

Solution

$$KE=eV=\dfrac{1}{2}m_eV^2$$
and $$R=\dfrac{m_eV}{eB}$$
$$=\dfrac{m_e}{eB}\times\sqrt{\dfrac{eV\times 2}{m_e}}$$
$$=\sqrt{\dfrac{2Vm_e}{eB^2}}$$
Question 8
1 / -0

A free charged particle moves through a magnetic field. The particle may undergo a change in

A
speed

B
energy

C
direction of motion

D
magnitude of the velocity

Solution

Hint:
When any charge particle with charge $$q$$ is moving with velocity $$v$$ in magnetic field $$B$$. A force acts on it which is given by,
$$F = qvB sin\theta$$
where, $$ \theta$$ is the angle between direction of velocity of charge and Magnetic Field.

Correct Option is C.

Explanation for correct answer:
$$ \bullet $$ Whenever any charge $$q$$ is moving with velocity $$v$$ in an magnetic field $$B$$ , a force acts on it, which is given by
$$F = qvB sin\theta$$
or,
$$F=q(v \times B)$$
Note that, the force is perpendicular to $$v$$ and $$B$$.

$$ \bullet $$ The force is perpendicular to the direction of motion of the charge. Thus it do not accelerate the charge, in other words it do not changes the magnitude of velocity of charge, it just changes the direction of motion of the charge.
Question 9
1 / -0

The mono energetic beams of electrons moving along +y direction enter a region of uniform electric and magnetic fields. If the beam goes straight through these simultaneously, then field B and E are directed possibly along

A
-y axis and -z axis

B
+z axis and -x axis

C
+ x axis and - x axis

D
-x axis and -y axis

Solution

$$\overrightarrow{V} =V({\widehat{j}})$$
Hence , B and E should be perpendicular to each other and also perpendicular to the velocity . Hence , B and E should be in z - axis and x axis
If B is towards + z axis , Force = q(V \times B ) is in + x direction . To balance this , E has to be in -x direction
Question 10
1 / -0

A charged particle enters into a uniform magnetic field the parameter that remains constant is

A
velocity

B
momentum

C
kinetic energy

D
angular velocity

Solution

Since the force on the charge is always normal to the velocity of the charge so, field does not work on the particle and the magnitude of velocity remains same. So, kinetic energy is same, whereas velocity changes direction. So, velocity, momentum and angular velocity changes.

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Moving Charges and Magnetism Test - 33

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Moving Charges and Magnetism Test - 33

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Question 1

the wire oscillates

the wire continues making contact

the wire breaks contact just when the current is passed

the mercury will expand by heating due to passage of current

Solution

Question 2

current through it

distance from it

its length

nature of meterial

Solution

Question 3

vertically upwards

vertically downwards

horizontal and perpendicular to the direction ofmotion of the electron beam

horizontal and parallel to the direction of motionof the electron beam

Solution

Question 4

electron

proton

both experience same force

it cannot be predicted.

Solution

Question 5

a straight line

an ellipse

a circle

a helix.

Solution

Question 6

remain unaffected

start moving in a circular path in $$Y-Z$$ plane

retard along $$X-$$axis

move along a helical path around $$X -$$ axis

Solution

Question 7

$$\sqrt{\dfrac{2eV}{m}}$$

$$\sqrt{\dfrac{2Vm}{eB^{2}}}$$

$$\sqrt{\dfrac{2Vm}{eB}}$$

$$\sqrt{\dfrac{2Vm}{e^{2}B}}$$

Solution

Question 8

speed

energy

direction of motion

magnitude of the velocity

Solution

Question 9

-y axis and -z axis

+z axis and -x axis

+ x axis and - x axis

-x axis and -y axis

Solution

Question 10

velocity

momentum

kinetic energy

angular velocity

Solution

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horizontal and perpendicular to the direction of

motion of the electron beam

horizontal and parallel to the direction of motion

of the electron beam