Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 29

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Question 1
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A charged particle is moving in a uniform magnetic field. Identify which of the following trajectory is not possible.

A

B

C

D

E

Solution

By using Fleming's left hand rule:
Trajectory $$A$$ is possible when-the magnetic field is in the direction of movement of charged particle, that is into the page. Because in this condition the magnetic force is zero as magnetic field and velocity are parallel to each other.east direction.
Trajectory $$B$$ is possible when-the magnetic field is in the direction of movement of charged particle, that is toward the west-direction. Because in this condition the magnetic force is zero as magnetic field and velocity are parallel to each other.
Trajectory $$C$$ is possible when-the magnetic field is in perpendicular to the movement of charged particle. Because in this condition the magnetic force will be perpendicular to both magnetic field and velocity of particle, according to Fleming's left hand rule.
Trajectory $$E$$ is possible when-the magnetic field is not perpendicular to the direction of magnetic field, in this condition the component of magnetic field perpendicular to velocity of particle causes the circular motion and other component causes the linear motion, resulting in a spiral motion.
Trajectory $$D$$ is not possible because it shows angular and linear trajectories separately whereas they must be combined as spiral trajectory.
Question 2
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The diagram above shows atomic particles moving through a magnetic field.
A beam of electrons is deflected in the magnetic field and strikes at the point $$P$$ as shown in the figure.
At which letter would a stream of neutrons strike the screen?

A
$$A$$

B
$$B$$

C
$$C$$

D
$$D$$

E
$$E$$

Solution

The force on a charge $$q$$ in a magnetic field $$B$$ moving with velocity $$v $$ is given by $$F=qvB\sin{\theta}$$ or $$\vec F=q\vec v\times\vec B$$ where $$\theta$$ is the angle between $$\vec v$$ and $$\vec B$$.
As neutron has no charge $$q=0$$, so it does not experience any magnetic force and will go straight to point B.
Question 3
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Which of the following relation represents Biot-Savart's law?

A
$$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ r } $$

B
$$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \hat { r } }{ { r }^{ 3 } } $$

C
$$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 3 } } $$

D
$$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 4 } } $$

Solution

If unit current is flowing through the conductor, then Biot-Savart's law is represented as
$$\vec { dB } =\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 3 } } $$
Question 4
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Which line shows the path of gamma radiation in a magnetic field?

A
A

B
B

C
C

D
D

E
E

Solution

Magnetic force on a charged particle is given by, $$F=qvB\sin\theta$$
We know that gamma rays are electromagnetic waves , having no charge. Therefore, they will not experience any force in a magnetic field, and will go straight in a magnetic field without any deflection i.e. path $$A$$.
Question 5
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Which of the following is based on mechanical effect of electric current?

A
AC Dynamo

B
DC Dynamo

C
AC or DC motor

D
Electric Geyser

Solution

Working of an electric motor is based on the mechanical effect of electric current. A conductor carrying a current is placed in a magnetic field experiences a mechanical force. In the motor, when a current is passed through a rectangular coil of wire placed in a magnetic field, the coil rotates continuously.
Question 6
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As shown above, a magnetic force of $$10^{-14}$$ N is experienced by a charge particle moving with speed of $$10^6$$ m/s in a uniform magnetic field B of magnitude $$10^{-2}$$ T. Calculate the magnitude of the charge.

A
$$10^{-22}$$ C

B
$$10^{-18}$$ C

C
$$10^{-10}$$ C

D
$$10^{-6}$$ C

E
$$10^{-2}$$ C

Solution

Given : $$F_m =10^{-14}$$ N ;$$B = 10^{-2}$$ T ;$$v = 10^6$$ m/s
Using $$F_m = |q|vB$$
$$\therefore$$ $$10^{-14} = |q| \times 10^6 \times 10^{-2} $$
$$\implies $$ $$|q| =10^{-18}$$ C
Question 7
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Directions For Questions

A small positive charge particle is moving in a straight line through the space. If the charge of the particle is $$q$$. Answer the following questions.

...view full instructions

A person who is at rest as the charge moves by him will measure.

A
A magnetic field and an electric field due to the moving charge.

B
A magnetic field, but not an electric field due to the moving charge.

C
An electric field, but not a magnetic field due to the moving charge.

D
Neither an electric field nor a magnetic field due to the moving charge.

E
A decrease in the amount of charge.

Solution

All charges produce an electric field whether they are moving or not, and magnetic fields are created by moving charges, so the person would measure both.
Question 8
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An electron of charge $$e$$ and mass $$m_e$$ moves in a circular path of radius $$r$$ in a uniform magnetic field $$B$$.
The Kinetic Energy of the electron can be described by the expression

A
$$\dfrac{1}{2}m_ev^2$$

B
$$evBr$$

C
$$\dfrac{1}{2}evBr$$

D
$$2evBr$$

E
$$\sqrt{\dfrac{1}{2}evBr}$$

Solution

Using $$eB r = mv$$ $$\implies v = \dfrac{eBr}{m}$$
$$\therefore$$ kinetic energy of the electron $$K.E = \dfrac{1}{2}m_ev^2$$ where $$v = \dfrac{eBr}{m}$$
Question 9
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A charged particle moves in a magnetic field. The only force influencing the particle is the force caused by the magnetic field.
During the particle's movement in the magnetic field, what will NOT change?

A
the particle's velocity

B
the particle's acceleration

C
the particle's speed

D
the particle's momentum

E
the particle's position

Solution

"From Newton's second law of motion, we know that when an object experiences a net force, the object will accelerate in the direction of that net force. Since the only force acting on the particle is the force exerted by the magnetic field, this magnetic field force must be the net force.
Forces by magnetic fields on moving charged particles always act in a direction that is perpendicular to the velocity of the particle, meaning that they will never change the speed of the particle.
Since in this case the particle is experiencing a net force, the particle must be accelerating.
All of the vector quantities here are constantly changing since the particle is constantly changing direction.
The velocity direction is changing, the acceleration direction is changing, the momentum direction is changing along with the velocity, and the position is obviously changing.
The particle's speed does not change therefore option "C" is right.
Question 10
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Which of the following particle would follow the path as shown in the above figure?

A
Proton

B
Electron

C
Neutron

D
x-ray

E
Photon of red ligth

Solution

Force on a moving charged particle through a magnetic field is given by $$\vec{F}=q(\vec{v}\times \vec{B})$$.
A positively charged particle would thus experience a force vertically downwards. Hence a vertically upwards force would be experienced by a negatively charged particle, that is, an electron to follow the given path.