Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 54

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Question 1
1 / -0

Identify in which of the following cases would the charge experience a magnetic force?
I. The stationary charge is placed inside the magnetic field
II. The charge is moving and its velocity is perpendicular to the magnetic field lines
III. The charge is moving and its velocity is parallel to the magnetic field lines

A
I only

B
I and II only

C
II and III only

D
II only

E
III only

Solution

Magnetic force $$F = qvB$$ $$sin\theta$$ where $$\theta$$ is the angle between $$v$$ and $$B$$
$$I$$ : $$v = 0$$ $$\implies F = 0$$
$$II$$ : $$\theta =90^0$$ $$\implies F = qvB$$ $$\times sin 90^o = qvB$$
$$III$$ : $$\theta =0^0$$ $$\implies F = qvB$$ $$\times sin 0^o = 0$$
Hence option D is correct.
Question 2
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A charged particle is moving in circular path of radius 100 meters in a uniform magnetic field. If mass of the charged particle is $$9.1$$ x $$10^{-31}$$ kg and its speed is 3 x $$10^7$$ m/s. Which of the following is the best estimate of the order of magnitude of the magnetic force needed to maintain this orbit?

A
$$10^{-22}$$

B
$$10^{-17}$$

C
$$10^{-10}$$

D
$$10^{-6}$$

E
$$10^{-2}$$

Solution

Given : $$r = 100$$ m; $$m = 9.1 \times 10^{-31} $$ kg; $$v = 3 \times 10^7$$ m/s
Magnetic force $$F = \dfrac{mv^2}{r} $$

$$= \dfrac{(9.1 \times 10^{-31}) \times (3\times 10^7)^2}{100} $$

$$ =8.2 \times 10^{-18} \approx 10^{-17}$$ N
Question 3
1 / -0

An electron moving with velocity 'v' enters a magnetic field as shown in the above figure.
Identify in which direction, the electron will experience a force ?

A
into the page.

B
out of the page.

C
toward the top of the page.

D
toward the bottom of the page.

E
to the left.

Solution

The component of the velocity that is perpendicular to the direction of the magnetic field influences the magnetic force acting on the charge.
Therefore from Fleming's-left hand rule the magnetic force for the given directions act into the plane of the page for a positive charge.
But the given charge in the question is negative, thus the magnetic force on this charge is acting out of the plane of the page.
Question 4
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A particle of mass m is moving in a circular path in a uniform magnetic field as shown above. What is the sign of charge on the particle?

A
Negative

B
Positive

C
Negative or Positive

D
Neutral

E
Cannot be determined

Solution

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Question 5
1 / -0

A proton of charge $$e$$ moving at speed $$v_0$$ is placed midway between two parallel wires a distance $$a$$ apart, each carrying current $$I$$ in opposite directions.
The force on the proton is:

A
$$0$$

B
$$ev_0\cfrac{_0I}{2\pi a}$$

C
$$ev_0\cfrac{_0I}{2\pi (2a)}$$

D
$$2ev_0\cfrac{_0I}{2\pi \frac{a}{2}}$$

E
Unable to be determined
Question 6
1 / -0

A proton of charge $$'e'$$ moving at speed $$'v_0\ '$$ is placed midway between two parallel wires $$a$$ distance a apart, each carrying current $$I$$ in the same direction.
The force on the proton is

A
$$0$$

B
$$ev_0\cfrac{_0I}{2\pi a}$$

C
$$ev_0\cfrac{_0I}{2\pi (2a)}$$

D
$$ev_0\cfrac{_0I}{2\pi \frac{a}{2}}$$

E
Unable to be determined

Solution

Magnetic field due to a straight wire at a distance d $$B =\dfrac{\mu_o I}{2\pi d}$$

Thus magnetic field at p due to wire 1 $$B_1 = \dfrac{\mu_o I}{2\pi (a/2)} = \dfrac{\mu_o I}{\pi a}$$ (into the paper)

Magnetic field at p due to wire 2 $$B_2 = \dfrac{\mu_o I}{2\pi (a/2)} =\dfrac{\mu_o I}{\pi a}$$ (out of the paper)
$$\therefore$$ Net magnetic field at p $$B_{net} =B_1 - B_2 = 0$$
$$\implies$$ Magnetic force experienced by proton $$F = ev_o B_{net} = 0$$ Newton
Question 7
1 / -0

A proton of charge $$e$$ moving at speed $$v_0$$ is placed midway between two parallel wires $$a$$ distance a apart, each carrying current $$I$$ in the same direction.
The force on the proton is:

A
$$0$$

B
$$ev_0\cfrac{I_0}{2\pi a}$$

C
$$ev_0\cfrac{I_0}{2\pi (2a)}$$

D
$$ev_0\cfrac{I_0}{2\pi \frac{a}{2}}$$

E
Unable to be determined

Solution

The force $$F$$ on a proton $$e$$ moving with velocity $$v_{0}$$ in a magnetic field $$B$$ is given by,

$$F=ev_{0}B\sin\theta$$

where $$\theta=$$ angle between v and B,

now magnetic field due to a wire carrying current I, at a distance a/2(midpoint) is,

$$B=\dfrac{\mu_{0}.I}{2\pi(a/2)}$$

magnetic field due to another wire carrying current I , at a distance a/2(midpoint) is,

$$B'=\dfrac{\mu_{0}.I}{2\pi(a/2)}$$ ,

therefore magnitude of force on proton is same due to magnetic fields by both wires as $$B=B'$$ if we apply right hand thumb rule to find the direction of magnetic fields, we get that both fields are opposite in direction,

now by Fleming's left-hand thumb rule the direction of magnetic forces due to both wires, on proton, are opposite to each other therefore being equal in magnitude but opposite in direction, they will cancel each other and proton will experience no force.
Question 8
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A proton of charge $$e$$ and mass $$m_p$$ moves in a circular path of radius $$r$$ in a uniform magnetic field $$B$$.
The momentum of the proton can be described by the expression:

A
$$eBr$$

B
$$2eBr$$

C
$$eBr^2$$

D
$$eBrm_p$$

E
$$2eBrm_p$$

Solution

Momentum of the proton $$P = m_pv$$
Using $$eB r = m_pv$$
$$\implies$$ $$P = eBr$$
Question 9
1 / -0

Two wires are pictured below, both carrying current toward the east. What is the direction of the force exerted by wire 2 on wire 1?

A
north

B
south

C
up

D
down

E
east

Solution

When the current will flow in a same direction through the two parallel wires, the force between them will be attractive and when the current in opposite direction, the force between them will be repulsive.
As force is attractive here, so the force on wire 1 due to 2 will be acted northward.
Question 10
1 / -0

The diagram above shows a negatively charged ball moving to the right below a wire carrying conventional current to the right. Both are in the plane of the screen.
What is the direction of the force on the negatively charged ball at the place it is pictured?

A
up toward the top of the screen

B
down toward the bottom of the screen

C
toward the left side of the screen

D
toward the right side of the screen

E
The force is zero, since the negative charge is moving parallel to the current.

Solution

The direction of magnetic field lines in the region right to the current carrying wire where the negative charge is moving will be perpendicularly inward the page according to the right thumb rule .
The direction of movement of negative charge can be taken as the direction opposite to the direction of current . As the negative charge is moving toward right therefore the direction of current will be toward ,right , and magnetic field is perpendicular inward the paper so according to Fleming's left hand rule , force on positive charge will be down toward the bottom of the page .