Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 76

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Question 1
1 / -0

A neutron, a proton, an electron and an $$\alpha$$ particle enters a uniform magnetic field with equal velocities. The field is directed along the inward normal to the plane of the paper. Which of these tracks followed are by electron and $$\alpha$$ particle ?

A
A

B
B

C
C

D
D

Solution
Question 2
1 / -0

An alpha particle and proton have same velocity when enter uniform magnetic field. The period of rotation of proton will be

A
Double of that of $$\alpha$$ particle

B
Four times that of $$\alpha$$ particle

C
One half times that of $$\alpha$$ particle

D
Same as that of $$\alpha$$ particle.

Solution
Question 3
1 / -0

A charged particle moves through a magnetic field perpendicular to its direction, then

A
the momentum changes but the K.E is constant

B
both momentum and K.E of the particle are not constant

C
both momentum and K.E of the particle are constant

D
K.E changes but the momentum is constant

Solution

Momentum changes but K.E remains constant because particles gets deflected and moved in a curved path so the velocity changes as the direction of the speed changes hence the movement on the body also changes while K.E unchanged.
Question 4
1 / -0

If the radius of the circle is 0.5 m and the magnitude of the magnetic field is $$1.2 W bm^{-2}$$, find the frequency and the kinetic energy of the proton. Charge of the proton =$$ 1.60\times 10^{-19}C.$$ Mass of the proton = $$1.67\times 10^{-27}kg$$.

A
$$1.83\times 10^7Hz, 2.76 \times 10^{-12}J$$

B
$$1.83\times 10^7Hz, 3.76 \times 10^{-12}J$$

C
$$2.83\times 10^7Hz, 2.76 \times 10^{-12}J$$

D
$$2.83\times 10^7Hz, 3.76 \times 10^{-12}J$$

Solution
Question 5
1 / -0

An electron moves straight inside a charged parallel plate capacitor of uniform surface charged density $$\sigma$$. The length of capacitor plates is $$l$$. The space between the plates is filled with constant magnetic field of induction $$\overrightarrow {B} $$. The time of straight line motion of the electron in the capacitor is:

A
$$\dfrac{e \sigma}{\epsilon_0l B}$$

B
$$\dfrac{\epsilon_0lB}{\sigma}$$

C
$$\dfrac{e \sigma }{\epsilon_0B}$$

D
$$\dfrac{\epsilon_0B}{e\sigma}$$

Solution

The net electric field

$$E= \overrightarrow {E}_1+ \overrightarrow {E}_2$$

$$=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}= \dfrac{\sigma}{2\epsilon_0}$$

The net force acting on the electron is zero because it moves with constant velocity, due to its motion on straight line.

$$\Rightarrow \overrightarrow {F}_{net}= \overrightarrow {F}_{e}+ \overrightarrow {F}_{m}=0$$

$$|\overrightarrow {F}_{e}|=| \overrightarrow {F}_m|$$

$$\Rightarrow eE= eVB$$

$$\Rightarrow V= \dfrac{E}{B} =\dfrac{\sigma}{\epsilon_0B}$$

$$\therefore $$ The time of motion in side the capacitor $$= t= \dfrac{l}{V}= \dfrac{\epsilon_0 l B}{\sigma}$$.
Question 6
1 / -0

A charged particle $$q$$ enters a region of uniform magnetic field (out of the page) end is deflected a distance $$d$$ after travelling a horizontal distance $$a$$. The magnitude of the momentum of the particle is:

A
$$\dfrac{qB}{2}\left[\dfrac{a^{2}}{d}+d\right]$$

B
$$\dfrac{qB}{2}$$

C
Zero

D
Not possible to be determined as it keeps changing.

Solution

The vector product between $$v$$ and $$B$$ points upwards in the figure thus indicating that the charge of the particle is positive.
The Force acting on the moving charge $$F = qvB $$.
As a result of the magnetic force, the charged particle will follow a spherical trajectory.
The centripetal force $${F_c} = \dfrac{{m{v^2}}}{r}$$.
Force acting on the moving charge $$F = F_c$$,
$$\ qvB = \dfrac{{m{v^2}}}{r}$$
$$r = \dfrac{{mv}}{{qB}} = \dfrac{p}{{qB}}$$,where $$p$$ is the momentum of the particle.
$$\begin{array}{l} { r^{ 2 } }={ d^{ 2 } }+{ \left( { r-a } \right) ^{ 2 } } \\ =\dfrac { { { d^{ 2 } }+{ a^{ 2 } } } }{ { 2d } } \\ =\dfrac { 1 }{ 2 } \left( { d+\dfrac { { { a^{ 2 } } } }{ d } } \right) \end{array}$$
So,
$$\begin{array}{l} p=qBr \\ =\dfrac { { qB } }{ 2 } \left( { d+\dfrac { { { a^{ 2 } } } }{ d } } \right) \end{array}$$
Question 7
1 / -0

Consider a region where both uniform electric and magnetic fields E and B are present both along the z-axis. A positively charged particle of charge and mass is released from the origin with an initial velocity $${{\text{V}}_e}\hat i$$. Which of the following option(s) are correct?

A
(A)The y coordinate of the particle at time $${\text{t}} = \frac{{\pi {\text{M}}}}{{{\text{qB}}}}{\text{ is}}\frac{{ - 2{\text{mv}}}}{{{\text{qB}}}}$$

B
(B)The distance between two consecutive point on the z-axis where the particle touches the Z-axis is an odd multiple of a constant distance.

C
(C)The distance between two consecutive point on the z-axis where the particle touches the Z-axis is an even multiple of a constant distance.

D
(D)The time after which the particle touches the z-axis is $$\frac{{2\pi {\text{m}}}}{{{\text{qB}}}}$$

Solution
Question 8
1 / -0

H$$^ + $$, He$$^ + $$ and O$$^ + $$$$^ + $$ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. the masses of H$$^ + $$, He$$^ + $$ and O$$^ + $$$$^ + $$ are 1 a.m.u., 4 a.m.u. and 16 a.m.u. respectively. Then

A
H$$
^ +
$$
will be deflected most

B
O$$
^ +
$$
$$
^ +
$$
will be deflected most

C
He$$
^ +
$$
and O$$
^ +
$$
$$
^ +
$$
will be deflected unequally

D
All will be deflected equally

Solution
Question 9
1 / -0

Two particle, A and B having equal mass from being accelerated through the same potential difference enter a region of uniform magnetic field and particle describe a circular path of radii $$R_1$$ and $$R_2$$ respectively. The ratio of the charge on A and B is:

A
$$\lgroup \frac{R_2}{R_1} \rgroup^{\frac{1}{2}}$$

B
$$\frac{R_2}{R_1}$$

C
$$\lgroup \frac{R_2}{R_1} \rgroup^2$$

D
$$\lgroup \frac{R_1}{R_2} \rgroup^2$$

Solution
Question 10
1 / -0

An $$\alpha $$ - particle crosses a space without any deflection. If electric field E = 8 x $$10^{6} $$ V/m and magnetic field is B = 1.6 T, the velocity of particle is

A
(a) 2.5 x $$10^{6} $$ m/s

B
(b) 5 x $$10^{6} $$ m/s

C
(c) 8 x $$10^{6} $$ m/s

D
(d) 5 x $$10^{7} $$ m/s

Solution