Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

If we want to maximise the magnitude of the magnetic force acting on a charged particle moving in a magnetic field. By which of the following statements we can maximise it.
I. Maximize the strength of the magnetic field
II. Minimize the velocity of particle.
III. Ensure that the particle is moving in the same direction as the magnetic field lines

A
I only

B
I and II only

C
I and III only

D
II and III only

E
I, II, and III

Solution

Magnetic force, $$F = q vB $$ $$sin\theta$$ where $$\theta$$ is the angle between $$v$$ and $$B$$
Thus force can be maximizes either by maximizing $$v$$ or $$B$$ or $$\theta = 90^o$$
Thus only statement $$I$$ is true and hence option A is correct.
Question 2
1 / -0

A charged particle is moving in a circular orbit in a magnetic field. If the strength of the magnetic field becomes four times, Find out the change in radius of the particle's orbit.

A
It is quartered

B
It is halved

C
It is unchanged

D
It is doubled

E
It is quadrupled

Solution

Using $$Bqr = mv$$ $$\implies r = \dfrac{mv}{Bq}$$
Now the magnetic field is increased to 4 times i.e $$B' = 4B$$
$$\therefore$$ New radius $$r' = \dfrac{mv}{B'q} = \dfrac{mv}{(4B)q} = \dfrac{r}{4}$$
Thus the radius is quartered.
Question 3
1 / -0

A particle with velocity $$2\times10^4m/s$$ enters the uniform magnetic field perpendicularly. Calculate the magnitude of the magnetic force on this particle if charge of particle is -0.04 C and magnetic field of strength is B= 0.5 T.

A
4 N

B
8 N

C
40 N

D
80 N

E
400 N

Solution

Magnetic force $$F = q (v \times B)$$
As $$v \perp B$$ $$\implies |F| =|q|vB$$
$$\therefore$$ $$|F| = 0.04 \times (2\times 10^4) \times 0.5 = 400$$ $$N$$
Question 4
1 / -0

A positively charged particle is moving in uniform circular path due to magnetic force in uniform magnetic field.Which of the following changes could cause the radius of the circular path to decrease?

A
Increase the mass of the particle

B
Increase the speed of the particle

C
Decrease the charge of the particle

D
Decrease the strength of the magnetic field

E
None of the above

Solution

Let the charge and mass of the particle be $$q$$ and $$m$$ respectively moving in a circular path in the region of uniform magnetic field $$B$$
Using $$Bqr = mv$$ $$\implies r = \dfrac{mv}{Bq}$$
Thus either increasing the mass or speed of the particle or decreasing the charge of particle or strength of magnetic field results in the increase of the radius of circular orbit.
Hence option E is correct.
Question 5
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A negatively charged particle is moving in a circular path in the clockwise direction in a uniform magnetic field, parallel to the plane of the page. Find out the direction in which magnetic field lines point?

A
Out of the page

B
Into the page

C
To the left

D
To the right

E
In a clockwise pattern parallel to the plane of the page

Solution

As particle is in circular path, therefore magnetic field will be perpendicular to velocity of particle. For circular motion centripetal force must be directed toward centre, here magnetic force is acting as centripetal force so the direction of force is toward center at any instant. Let the negative (current will be in opposite direction) particle is moving toward right (due to clockwise direction) at the top of circle, force will be vertically downward (toward center of circle) therefore by Fleming's left hand rule, magnetic field is into the page.
Question 6
1 / -0

A charged particle is moving in a circular orbit in a magnetic field. If the strength of the magnetic field doubles, how does the radius of the particles orbit change?

A
It is quartered

B
It is halved

C
It is unchanged

D
It is doubled

E
It is quadrupled

Solution

Initial radius of the orbit $$r = \dfrac{mv}{Bq}$$

New strength of magnetic field $$B' = 2B$$

Thus radius of new orbit $$r' = \dfrac{mv}{(2B) q} = \dfrac{r}{2}$$
Question 7
1 / -0

Two wires carry current in opposite directions. Which of the following graphs represents the magnetic force acting on each wire?

A

B

C

D

E
There is no net force acting on either wire

Solution

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Question 8
1 / -0

Which of the following is not a possible trajectory of a charged particle in a uniform magnetic field?

A

B

C

D

E

Solution
Question 9
1 / -0

What should one do to maximize the magnitude of the magnetic force acting on a charged particle moving in a magnetic field?
I. Maximize the strength of the magnetic field
II. Minimize the particles velocity
III. Ensure that the particle is moving in the same direction as the magnetic field lines.

A
I only

B
I and II only

C
I and III only

D
II and III only

E
I, II, and III

Solution

Magnitude of magnetic force $$F$$ ,acting on a charged particle $$q$$ moving with velocity $$v$$ in a magnetic field of strength $$B$$ is given by ,
$$F=qvB\sin\theta$$ ,
where $$\theta=$$ angle between $$v$$ and $$B$$ ,
it is clear from this relation force will be maximized , when magnetic field strength $$B$$ is maximum .
if we minimize the velocity $$v$$ , force $$F$$ will also be minimized , and if particle moves in the direction of field then $$\theta=0$$ which gives
$$F=qvB\sin0=0$$
Question 10
1 / -0

Two positively charged particles, one twice as massive as the other, are moving in the same circular orbit in a magnetic field. Which law explains to us why the less massive particle moves at twice the speed of the more massive particle?

A
Coulomb's Law

B
Conservation of angular momentum

C
Hooke's Law

D
The ideal gas law

E
Heisenberg's uncertainty principle

Solution

As there is no external torque on the given system of charges , therefore total angular momentum $$L $$ of the system remains constant according to law of Conservation of angular momentum i.e. ,
$$L=I_{1}\omega_{1}+I_{2}\omega_{2}$$ , ................$$eq1$$
where $$I_{1}=m_{1}r^{2}$$ (moment of inertia of first particle )
$$I_{2}=m_{2}r^{2}$$ (moment of inertia of second particle )
let first particle is massive than second, given that the mass of first particle is twice of the second,
$$m_{1}=2m_{2}$$
therefore $$I_{1}=2I_{2}$$ ,
if we use this relation in $$eq1$$, we can see that ,that the value of $$L$$ will be constant only when
$$\omega_{2}=2\omega_{1}$$ ,
or $$v_{2}=2v_{1}$$ (given , $$r$$ is same for both particle)