Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle

A
depends on both v and R

B
depends on v and not on R

C
depends on R and not on v

D
is independent of both v and R

Solution

The centripetal force of the charged particle is provided by the magnetic force acting on the particle.

Hence $$qvB=\dfrac{mv^2}{r}$$

$$\implies \dfrac{r}{v}=\dfrac{m}{qB}$$

Hence time period of the rotation of particle is

$$T=\dfrac{2\pi r}{v}=\dfrac{2\pi m}{qB}$$

Hence it is independent of both $$v$$ and $$r$$.
Question 2
1 / -0

An electron has a circular path of radius $$0.01 m$$ in a perpendicular magnetic induction $$10^{-3} T$$. The speed of the electron is nearly

A
$$7.04 \times 10^{6}\ {ms}^{-1}$$

B
$$1.76 \times 10^{6}\ {ms}^{-1}$$

C
$$3.52 \times 10^{6}\ {ms}^{-1}$$

D
$$1.76 \times 10^{4}\ {ms}^{-1}$$

Solution

If a charge moves into a magnetic field with direction perpendicular to the field, it will follow a circular path. The magnetic force, being perpendicular to the velocity, provides the centripetal force.(as shown in figure)

Now, $$F=qvB=\dfrac{mv^2}{r}$$

$$v=\dfrac{rqB}{m}$$

$$v=\dfrac{0.01\times1.6021\times{10^-19}\times{10^-3}}{9.1093\times{10^-31}}=1.76\times{10^6}$$ m/sec
Question 3
1 / -0

Figure shows the path of an electron in a region of uniform magnetic field. The path consists of two straight sections, each between a pair of uniformly charged plates and two half-circles. The plates are named 1,2,3 and 4. Then

A
1 and 3 at higher (positive) potential and 2 and 4 at lower (negative) potential

B
1 and 3 at lower potential and 2 and 4 at higher potential

C
1 and 4 at higher potential and 2 and 3 at lower potential

D
1 and 4 at lower potential and 2 and 3 at higher potential

Solution

as shown in figure
$$q=-ve (electron),\vec{\vartheta }=\vartheta (-\hat{i})$$ and
$$\vec{F}=F(-\hat{j})$$
$$\therefore $$ using $$\vec{F}=q(\vec{\vartheta }*\vec{B})$$
$$F(-\hat{j})=-e\vartheta (-\hat{i}*\vec{B})$$
$$F\hat{j}=-e\vartheta (\hat{i}*\vec{B})$$
Hence direction of $$\vec{B}$$ must be along Z- axis (along $$\hat{k}$$)
In case of upper plates, to balance downward force by $$\vec{B},\vec{E}$$ must apply force upwards and hence plate 1 should be at higher, potential compared to 2.
Similarly in lower region, force acting by $$\vec{B}$$ is upward and hence to balance it plate 4 should be at higher potential compared to plate 3.
Question 4
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Directions For Questions

As a charged particle $$q$$ moving with a velocity $$\vec { v } $$ enters a uniform magnetic field $$\vec { B } $$, it experiences a force $$\vec { f } =q(\vec { v } \times \vec { B } )$$.For $$\theta={0}^{o}$$ or $${180}^{o}$$, $$\theta$$ being the angle between $$\vec { v } $$ and $$\vec { B } $$, force experienced is zero and the particle passes undeflected. For $$\theta={90}^{o}$$, the particle moves along a circular arc and the magnetic force $$(qvB)$$ provieds the necessary centripetal force $$(\cfrac{m{v}^{2}}{r})$$. For other values of $$\theta$$ ($$\theta\ne {0}^{o}, {180}^{o}, {90}^{o}$$), the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions. Suppose a particle, that carries a charge of magnitude $$q$$ and has a mass $$4\times {10}^{-15}kg$$, is moving in a region containing a uniform magnetic field $$\vec { B } =-0.4\hat { k } T$$. At a certain instant, velocity of the particle is $$\hat { k } =(8\hat { i } -6\hat { j } +4\hat { k } )\times {10}^{6}m/s$$ and force acting on it has a magnitude $$1.6N$$

...view full instructions

Which of the three components of acceleration have non-zero values?

A
$$x$$ and $$y$$

B
$$y$$ and $$z$$

C
$$z$$ and $$x$$

D
$$x, y$$ and $$z$$

Solution

Force on a moving charged particle placed in an external magnetic field is

$$\vec{F}=q(\vec{v}\times \vec{B})$$

$$=q[(8\hat{i}-6\hat{j}+4\hat{k})\times (-0.4\hat{k})]$$

The z-component here vanishes.
Hence force and hence acceleration has non-zero values in $$x$$ and $$y$$ directions only.
Question 5
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Two charges of same magnitude move in two circles of radii $$R_{1} = R$$ and $$R_{2} = 2R$$ in a region of constant uniform magnetic field $$B_{0}$$.
The work $$W_1$$ and $$W_2$$ done by the magnetic field in the Two cases, respectively are such that

A
$$W_{1} = W_{2} = 0$$

B
$$W_{1} > W_{2}$$

C
$$W_{1} = W_{2} \neq 0$$

D
$$W_{1} < W_{2}$$

Solution

As given that two circle have radii $${R}_{1}$$ and $${R}_{2}$$, therefor the force will act towards the center of both the circle.
Now work done is given by $$W=F{ \cos { \theta } }$$
Where,$${\theta}$$=$${90}^{0}$$
hence work done $${W}_{1}={W}_{2}=0$$
Question 6
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When the free ends of a tester are dipped into a solution, the magnetic needle shows deflection.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

The deflection in the compass needle shows that current is flowing through the wounded wire and hence, through the circuit. The circuit is complete since free ends of the tester are dipped in a solution.The solution is certainly a conducting solution. That is the reason why the compass needle shows a deflection.
Question 7
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Directions For Questions

A uniform magnetic field of $$30\times {10}^{6}T$$ exits in the $$+X$$ direction. A particle of charge $$+e$$ and mass $$1.67\times {10}^{-27}kg$$ is projected through the field in the $$+Y$$ direction with a speed of $$4.8\times {10}^{6}m/s$$

...view full instructions

If the particle were negatively charged then what will be the magnitude of force on the charged particle?

A
$$23.04\times {10}^{-6}N$$

B
$$230.4\times {10}^{-5}N$$

C
zero

D
None of these

Solution

Force acting on a charged particle moving in a magnetic field is

$$\vec{F}=q(\vec{v}\times \vec{B})$$

$$\implies \left|\vec{F}\right|=\left|-e\right|vB=evB$$

$$=1.6\times 10^{-19}\times 4.8\times 10^6\times 30\times 10^6N$$

$$=23.04\times 10^{-6}N$$.
Question 8
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Directions For Questions

A uniform magnetic field of $$30\times {10}^{6}T$$ exits in the $$+X$$ direction. A particle of charge $$+e$$ and mass $$1.67\times {10}^{-27}kg$$ is projected through the field in the $$+Y$$ direction with a speed of $$4.8\times {10}^{6}m/s$$

...view full instructions

The force on the charged particle in magnitude is

A
$$23.04\times {10}^{-6}N$$

B
$$230.4\times {10}^{-5}N$$

C
$$0$$

D
None of these

Solution

Force acting on a charged particle moving in a magnetic field is

$$\vec{F}=q(\vec{v}\times \vec{B})$$

$$\implies \left|\vec{F}\right|=qvB$$

$$=1.6\times 10^{-19}\times 4.8\times 10^6\times 30\times 10^6N$$

$$=23.04\times 10^{-6}N$$.
Question 9
1 / -0

A proton is moving with velocity $${10}^{4}m/s$$ parallel to the magentic field of intensity 5 tesla.The force on the proton is

A
$$8\times {10}^{-15}N$$

B
$${10}^{4}N$$

C
$$1.6\times {10}^{-19}N$$

D
Zero

Solution

Answer is D.

The magnitude and direction of F depend on the velocity of the particle and on the magnitude and direction of the magnetic field B.
When a charged particle moves parallel to the magnetic field vector, the magnetic force acting on the particle is zero.
When the particles velocity vector makes any angle with the magneticfield, the magnetic force acts in a direction perpendicular to both v and B; that is, F is perpendicular to the plane formed by v and B.
The magnitude of the magnetic force is
$$F=qvBsin\theta $$
where $$\theta $$ is the smaller angle between v and B. From this expression, we see that F is zero when v is parallel or antiparallel to B or 180) and maximum when v is perpendicular to B.
In this case, as the proton moves parallel to the magnetic field, the force is zero.
Question 10
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Two charges of same magnitude move in two circlesof radii $$R_1 = R$$ and $$R_2 = 2R$$ in a region of constant uniform magnetic field $$B_0$$. The work $$W_1$$, and $$W_2$$ done by the magnetic field in the Two cases, respectively are such that:

A
$$W_1 = W_2 = 0$$

B
$$W_1 > W_2$$

C
$$W_1 = W_2 \neq 0$$

D
$$W_1 < W_2$$

Solution

Answer is A.

In the presence of magnetic field, a moving charge experiences centripetal force given as $${ F }_{ B }=qvBsin\theta $$, which acts perpendicular to the velocity vector v.
Since the trajectory of such charge particle is circular, at any instant velocity vector acts along the tangent to the circle.$${ F }_{ B }$$ is perpendicular to v.
So work done by magnetic Lorentz force FB (due to B) will always be zero.
Hence, correct option is A, W1 = W2 = 0.

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Moving Charges and Magnetism Test - 49

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Moving Charges and Magnetism Test - 49

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Question 1

depends on both v and R

depends on v and not on R

depends on R and not on v

is independent of both v and R

Solution

Question 2

$$7.04 \times 10^{6}\ {ms}^{-1}$$

$$1.76 \times 10^{6}\ {ms}^{-1}$$

$$3.52 \times 10^{6}\ {ms}^{-1}$$

$$1.76 \times 10^{4}\ {ms}^{-1}$$

Solution

Question 3

1 and 3 at higher (positive) potential and 2 and 4 at lower (negative) potential

1 and 3 at lower potential and 2 and 4 at higher potential

1 and 4 at higher potential and 2 and 3 at lower potential

1 and 4 at lower potential and 2 and 3 at higher potential

Solution

Question 4

$$x$$ and $$y$$

$$y$$ and $$z$$

$$z$$ and $$x$$

$$x, y$$ and $$z$$

Solution

Question 5

$$W_{1} = W_{2} = 0$$

$$W_{1} > W_{2}$$

$$W_{1} = W_{2} \neq 0$$

$$W_{1} < W_{2}$$

Solution

Question 6

True

False

Ambiguous

Data insufficient

Solution

Question 7

$$23.04\times {10}^{-6}N$$

$$230.4\times {10}^{-5}N$$

zero

None of these

Solution

Question 8

$$23.04\times {10}^{-6}N$$

$$230.4\times {10}^{-5}N$$

$$0$$

None of these

Solution

Question 9

$$8\times {10}^{-15}N$$

$${10}^{4}N$$

$$1.6\times {10}^{-19}N$$

Zero

Solution

Question 10

$$W_1 = W_2 = 0$$

$$W_1 > W_2$$

$$W_1 = W_2 \neq 0$$

$$W_1 < W_2$$

Solution

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