Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 43

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Question 1
1 / -0

Directions For Questions

Two long straight conducting wires with linear mass density $$'\lambda'$$ are suspended using cords so that both of them are horizontal and parallel to each other at a distance $$'d'$$ apart. The back ends of the wires are connected by a low resistance slack wire. A charged capacitor is now added across the wires such that its positive terminal is connected to far end and negative terminal is connected to near end as shown in figure. The capacitance of the capacitor is $$C$$. These connections are also made by slack wires. Assume that time to discharge is negligible. Initial charge on capacitor is $$Q_0$$.

...view full instructions

The wires 1 and 2

A
move apart

B
come closer

C
remain at their position

D
none of these

Solution

It is evident from the question that current in the two is in opposite direction so on using the formulas of current carrying straight conductor we can say that they will repel each other.
Question 2
1 / -0

A charge of $$0.04 C$$ is moving in a magnetic field of $$0.02 T$$ with a velocity $$10 m/s$$ in a direction making an angle $$30^o$$ with the direction of field. The force acting on it will be :

A
$$4\times 10^{-3} N$$

B
$$2\times 10^{-3} N$$

C
zero

D
$$8\times 10^{-3} N$$

Solution

Force on the moving charge is,
$$\vec F=q(\vec{v} \times \vec{B})= qvB \sin \theta= 0.04 \times 10 \times 0.02 \sin 30^o= 0.04 \times 10 \times 0.02 \times 0.5 = 4 \times 10^{-3} \:N$$
Question 3
1 / -0

When a conductor is rotated in a perpendicular magnetic field then, it's free electrons

A
move in the field direction.

B
move at right angles to field direction.

C
remain stationary.

D
move opposite to field direction.

Solution

Force on electron = $$ -q_e \vec v \times \vec B $$
the force is perpendicular to the magnetic field.
Question 4
1 / -0

Two wires carry currents of $$100\ A$$ and $$200\ A$$ respectively and they repel each other with a force of $$0.4\ N/m$$. The distance between them will be :

A
$$1\ m$$

B
$$1\ cm$$

C
$$50\ cm$$

D
$$25\ cm$$

Solution

Force due to two parallel current carrying conductor is,
$$F = \dfrac{\mu_0}{4\pi} \dfrac{2i_1i_2}{r}=10^{-7}\dfrac{ 2 \times 100 \times 200}{r}$$
$$\therefore 0.4 \times r = 4 \times 10^{-3} $$
$$\therefore r = 10^{-2}\: m = 1\: cm$$
Question 5
1 / -0

A neutron, a proton, an electron and an $$\alpha-$$particle enter a region of uniform magnetic field with equal velocities. The magnetic field is perpendicular to the plane of paper and directed into it. The tracks of particles are labeled in the figure. The neutron follows the track

A
A

B
B

C
C

D
D

Solution

Neutron does not carry any charge.
Hence, there is no force on the neutron and it travels undeflected.
Question 6
1 / -0

An electron accelerated through a potential difference V passes through a uniform transverse magnetic field and experiences a force F. If the accelerating potential is increased to 2V, the electron in the same magnetic field will experience a force

A
F

B
F/2

C
$$\sqrt 2F$$

D
2F

Solution

$$F=Bqv$$
But $$\frac {1}{2}mv^2=eV$$ or $$v=\sqrt {\frac {2eV}{m}}$$
$$\therefore F=Bq\sqrt {\frac {2eV}{m}}$$
$$\Rightarrow F\propto \sqrt V$$ and
$$F\propto \sqrt {2V}$$
$$\frac {F'}{F}=\sqrt 2$$ or $$F'=\sqrt 2F$$
Question 7
1 / -0

An electron and a proton are injected into a uniform magnetic field perpendicular to it with the same momentum. If both particles are fired with same momentum into a transverse electric field, then

A
electron trajectory is less curved

B
proton trajectory is less curved

C
both trajctories are equally curved

D
both trajectories are straight lines

Solution

let, momentum of electron and proton be p.
Transverse electric force on particle provides the centripetal force to support the circular motion.
If r is radius of curvature, $$ F_e = F_c $$

i.e. $$qE = m \dfrac{v^2}{r} $$.

Solving the above equation we get r = $$ \dfrac {p^2}{mqE} $$

where, $$p = momentum = mv$$

Given $$q$$ ,$$p$$ and $$E$$ are constant. So, we can say that $$ r \propto \dfrac{1}{m} $$

Radius of curvature is higher for a lighter mass particle which is an electron and vice versa for proton.
Question 8
1 / -0

An electron is moving along positive $$x-$$axis. To get it moving on an anticlockwise circular path in $$x-y$$ plane, a magnetic field is applied

A
along positive $$y-$$axis

B
along positive $$z-$$axis

C
along negative $$y-$$axis

D
along negative $$z-$$axis

Solution

For the particle to move anti-clockwise, force should be in $$\hat{j}$$ direction.

$$\vec{F} = -e(\vec{v} \times \vec{B})=-e( v\hat{i} \times \vec{B})$$

$$ \Rightarrow \vec{B}= B\hat{k}$$ since LHS is in $$\hat{j}$$ direction.
Question 9
1 / -0

A conducting rod PQ is moving parallel to X-axis in a uniform magnetic filed directed in positive Y-direction. The end P of the rod will become

A
negative

B
positive

C
neutral

D
sometimes negative

Solution

Force of positive charges in rod = $$ q \vec v \times \vec B $$
As $$v$$ is along x axis and $$B$$ is along y axis the force will be along the z axis. Therefore the positive charges will move towards the end P and it will become positive charged.
Question 10
1 / -0

An electron and a proton are injected into a uniform magnetic field perpendicular to it with the same momentum. If the two particles are injected into a uniform transverse electric field with same kinetic energy, then

A
electron trajectory is more curved

B
proton trajectory is more curved

C
both trajectories are equally curved

D
both trajectories are straight lines

Solution

Let, Kinetic energies of electron and proton be K.
Transverse electric force on particle provides the centripetal force to support the circular motion in this case.
If r is radius of curvature, $$ F_e = F_c $$

i.e. qE = m $$\dfrac{v^2}{r} $$.

Solving the above equation we get r = $$ \dfrac {2K}{qE} $$ where K = Kinetic Energy = $$\dfrac {mv^2}{2}$$

Here, q,E and K are constant. So, radius of curvature is same for both of them.

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Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 43

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Moving Charges and Magnetism Test - 43

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SHARING IS CARING

Question 1

move apart

come closer

remain at their position

none of these

Solution

Question 2

$$4\times 10^{-3} N$$

$$2\times 10^{-3} N$$

zero

$$8\times 10^{-3} N$$

Solution

Question 3

move in the field direction.

move at right angles to field direction.

remain stationary.

move opposite to field direction.

Solution

Question 4

$$1\ m$$

$$1\ cm$$

$$50\ cm$$

$$25\ cm$$

Solution

Question 5

A

B

C

D

Solution

Question 6

F

F/2

$$\sqrt 2F$$

2F

Solution

Question 7

electron trajectory is less curved

proton trajectory is less curved

both trajctories are equally curved

both trajectories are straight lines

Solution

Question 8

along positive $$y-$$axis

along positive $$z-$$axis

along negative $$y-$$axis

along negative $$z-$$axis

Solution

Question 9

negative

positive

neutral

sometimes negative

Solution

Question 10

electron trajectory is more curved

proton trajectory is more curved

both trajectories are equally curved

both trajectories are straight lines

Solution

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