Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m, in a plane perpendicular to magnetic field $$\vec B$$. The kinetic energy of a proton that discribes a circular orbit of radius 0.5 m in the same plane with the same magnetic field $$\vec B$$ is

A
200 keV

B
50 keV

C
100 keV

D
25 keV

Solution

$$r_{deutron} = \displaystyle \dfrac{\sqrt{2 m_d E_d}}{Bq} ; r_{proton} = \dfrac{\sqrt{2m_p E_p}}{Bq}$$
For same radius, B and q:
$$m_p E_p = m_d E_d \\$$
$$ \Rightarrow E_p = \dfrac{m_d}{m_p} E_d = \dfrac{2}{1} \times 50 = 100 keV$$
Question 2
1 / -0

Ampere's circuital law holds good for:

A
conduction current only

B
displacement current only

C
both conduction current and displacement current

D
none of these

Solution

In the steady state $$\oint{B.dl}=\mu_0(I)$$,
where $$I$$ is conduction current.
In non steady state $$\oint{B.dl}=\mu_ 0(I+I_d)$$,
where $$I_d$$ is displacement current.
Question 3
1 / -0

A charge $$q=-4\mu C$$ has an instantaneous velocity $$\vec{v}=(2\hat{i} -3\hat{j}+\hat{k})\times {10}^{6}m/s$$ in a uniform magnetic field $$\vec{B}=(2\hat{i} +5\hat{j}-3\hat{k})\times {10}^{-2}T$$.What is the force on the charge?

A
$$(-0.16\hat{i}-0.32\hat{j}-0.64\hat{k})N$$

B
$$(-0.16\hat{i}-0.12\hat{j}-0.44\hat{k})N$$

C
$$(-0.16\hat{i}-0.82\hat{j}-0.44\hat{k})N$$

D
$$(-0.12\hat{i}-0.12\hat{j}-0.64\hat{k})N$$

Solution

$$\overrightarrow { F } =q\overrightarrow { v } \times \overrightarrow { B } ,\\ where\quad q\quad has\quad a\quad positive\quad or\quad negative\quad sign,\\ on\quad substituting,\quad we\quad get\\ -0.16\hat { i } -0.32\hat { j } +-0.64\hat { k } \quad $$
Question 4
1 / -0

An electron travelling with a speed u along the positive x-axis enters into a region of magnetic field where $$B = - B_0 \widehat k (x > 0)$$. It comes out of the region with speed v then

A
$$v = u$$ at y > 0

B
$$v = u$$ at y < 0

C
v > u at y > 0

D
v > u at y < 0

Solution

From Lorents equation: $$F = -ev \widehat i \times B_0 (- \widehat k) = -eu B_0 \widehat j$$
Hence, it will complete a semicircular arc and comes out of the region at a position y, such that y < 0.
Question 5
1 / -0

A beam of electrons enters normally in a magnetic field. find the correct condition.

A
q(v⃗ ×B⃗ )

B
none

C
2 q(v⃗ ×B⃗ )

D
3 q(v⃗ ×B⃗ )

Solution

The electron entering the field normally experiences a force $$q(\vec{v}\times \vec{B})$$ in the vertically downward direction.
As the particle tends to bend downwards, the direction of velocity also changes and force again acts perpendicular to that direction. Thus the particle traces a circular path with force being directed to the center of the circle all the time.
Finally the particle leaves the magnetic field parallel to the direction of incidence.
Question 6
1 / -0

An electron moving with uniform velocity in x direction enters a region of uniform magnetic field along y-direction. Which of the following physical quantity(ies) is (are) non zero and remain constant ?
I. Velocity of the electron
II. Magnitude of the momentum of the electron.
III. Force On the electron.
IV. The kinetic energy of electron.

A
Only I and II.

B
Only III and IV.

C
All four

D
Only II and IV.

Solution

Velocity and force change due to change in direction but magnitude of P and KE of electron remain constant speed is constant.
Question 7
1 / -0

A closed curve encircles several conductors.The line integral $$\int{\vec{B}\cdot d\vec{l}}$$ around this curve is $$3.83\times {10}^{-7} T-m$$. What is the net current in the conductors?

A
$$0.1A$$

B
$$0.2A$$

C
$$0.3A$$

D
$$0.4A$$

Solution

From Ampere's law, for a closed loop carrying current
$$\int \vec{B}.d\vec{l}=\mu_0 i$$
$$i=\dfrac{\int \vec{B}.d\vec{l}}{\mu_0}$$
$$=\dfrac{3.83\times 10^{-7}}{\mu_0}$$
$$=0.3A$$
Question 8
1 / -0

What reading would you expect of a square-wave current, switching rapidly between $$+0.5A$$ and $$-0.5A$$, when passed through an ac ammeter?

A
$$0$$

B
$$0.5\space A$$

C
$$0.25\space A$$

D
$$1.0\space A$$

Solution

AC ammeter reads the rms value.

rms value of current=$$ \sqrt { \dfrac{ 0.5 T (+0.5)^2 + 0.5 T (-0.5)^2 }{T} }= \sqrt{ 0.5^2} = 0.5 A $$
Question 9
1 / -0

If the acceleration and velocity of a charged particle moving in a constant magnetic region is given by $$\vec{a}={a}_{1}\hat{i}+{a}_{2}\hat{k},\vec{v}={b}-{1}\hat{i}+{b}_{2}\hat{k}$$.[$${a}_{1},{a}_{2},{b}_{1}$$ and $${b}_{2}$$ are constants],then choose the wrong statement

A
Magnetic field may be along y-axis

B
$${a}_{1}{b}_{1}+{a}_{2}{b}_{2}=0$$

C
Magnetic field is along x-axis

D
Kinetic energy of particle is always constant

Solution

(a) $$\vec{B}\bot \vec{v}$$, so it may along y-axis
(b) $$\vec{F}\bot \vec{v} \therefore \vec{a}\bot \vec{v} =0$$
or $$\vec{a}\cdot \vec{v}=0$$ or $${a}_{1}{b}_{1}={a}_{2}{b}_{2}$$
(c) See the logic of option (a).
(d) Magnetic force cannot charge the kinetic energy of a particle.
Question 10
1 / -0

The figure shows the cross-section of two long coaxial tubes carrying equal currents $$I$$ in opposite directions. If $${B}_{1}$$ and $${B}_{2}$$ are magnetic fields at point 1 and 2, as shown in figure then

A
$${B}_{1}\neq 0;{B}_{2}=0$$

B
$${B}_{1}=0;{B}_{2}=0$$

C
$${B}_{1}\neq 0;{B}_{2}\neq 0$$

D
$${B}_{1}=0;{B}_{2}\neq 0$$

Solution

At distance r from centre

$$B=\displaystyle\dfrac{{\mu}_{o}}{2\pi }\dfrac{({i}_{in})}{r}$$ (From ampere's circle law)

For path-1, $${i}_{in}\neq 0 \therefore {B}_{1}\neq 0$$

For path-2, $${i}_{in} =0 \therefore {B}_{2} =0$$

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Moving Charges and Magnetism Test - 47

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Moving Charges and Magnetism Test - 47

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Question 1

200 keV

50 keV

100 keV

25 keV

Solution

Question 2

conduction current only

displacement current only

both conduction current and displacement current

none of these

Solution

Question 3

$$(-0.16\hat{i}-0.32\hat{j}-0.64\hat{k})N$$

$$(-0.16\hat{i}-0.12\hat{j}-0.44\hat{k})N$$

$$(-0.16\hat{i}-0.82\hat{j}-0.44\hat{k})N$$

$$(-0.12\hat{i}-0.12\hat{j}-0.64\hat{k})N$$

Solution

Question 4

$$v = u$$ at y > 0

$$v = u$$ at y < 0

v > u at y > 0

v > u at y < 0

Solution

Question 5

q(v⃗ ×B⃗ )

none

2 q(v⃗ ×B⃗ )

3 q(v⃗ ×B⃗ )

Solution

Question 6

Only I and II.

Only III and IV.

All four

Only II and IV.

Solution

Question 7

$$0.1A$$

$$0.2A$$

$$0.3A$$

$$0.4A$$

Solution

Question 8

$$0$$

$$0.5\space A$$

$$0.25\space A$$

$$1.0\space A$$

Solution

Question 9

Magnetic field may be along y-axis

$${a}_{1}{b}_{1}+{a}_{2}{b}_{2}=0$$

Magnetic field is along x-axis

Kinetic energy of particle is always constant

Solution

Question 10

$${B}_{1}\neq 0;{B}_{2}=0$$

$${B}_{1}=0;{B}_{2}=0$$

$${B}_{1}\neq 0;{B}_{2}\neq 0$$

$${B}_{1}=0;{B}_{2}\neq 0$$

Solution

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