Physics M.C.Q (1 Marks)

2. $\text{E}_\text{y},=\text{E}_\text{y}+\frac{\text{vB}_\text{z}}{\text{c}^2}$

3. $\text{B}'_\text{y}=\text{B}_\text{y}+\text{v}\text{E}_\text{z}$

Explanation:

$\text{qE}=\text{qvB}$

$\Rightarrow\text{e}=\text{vB}$ By dimensionally b & care wrong

$\Rightarrow\text{v}\text{E}=\text{v}^2\text{B}$

$\Rightarrow\text{B}=\frac{\text{vE}}{\text{v} ^2}$

1. $\overrightarrow{\text{E}}||\overrightarrow{\text{B}},\ \overrightarrow{\text{v}}||\overrightarrow{\text{E}}$

2. $\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}}$

Explanation:

$\Rightarrow\overrightarrow{\text{V}}\overrightarrow{\text{E}},\ \overrightarrow{\text{B}}\overrightarrow{\text{E}}$

In this case Magnetic force on the particle is zero & $\overrightarrow{\text{V}}$ is paralle to $\overrightarrow{\text{E}}.$ So charged particle goes undeflected in a region.

$\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}},$ But $\overrightarrow{\text{V}}$ is parallel to $\overrightarrow{\text{E}}.$

4. $\text{Zero}$

Explantion:

As the current is entering and exiting from two diametrically opposite points of a circular coil, the currents in the two semicircular sections are in the opposite direction.

The fields due to the two semi-circular sections at the centre is in the opposite direction.

Hence net feild is zero.

1. Electron

Explanation:

$\text{F}=\text{q}\text{VB}=\frac{\text{mv}^2}{\text{r}}$

$\text{r}=\frac{\text{mV}}{\text{qB}}$

charqe electron = charqe of proton = charqe of He⁺ = charqe of Li⁺ But mass of electron is Lowest.

$\therefore$ (the electron so smallest + circle made by)

3. Will be deviated by different angles and hence separate.

Explanation:

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

Charge proton is poritive = e

F_p = evB

Charge of electron is negative = -e

F_e = -evB

They will be deviated by different angles and Hence separate.

4. Li⁺⁺

Explanation:

$|\text{F}|=|\text{qVB}|$

charge of  Li⁺⁺ > charge of ( He⁺, proton, electron)

1. $\overrightarrow{\text{E}}$ must be perpendicular to $\overrightarrow{\text{B}}$

2. $\overrightarrow{\text{V}}$ must be perpendicular to $\overrightarrow{\text{E}}$

Explanation:

$\Rightarrow\text{E}\perp\overrightarrow{\text{B}}\ \&\ \overrightarrow{\text{V}}\perp\overrightarrow{\text{E}}$

2. The circle described by the singly-ionised charge will have a radius that is. double that of the other circle.

4. The two circles touch each other.

Explanation:

$\text{r}=\frac{\text{mv}}{\text{qB}}$

If charge of singly ionized = e

Then charge of doubly ionized = ze

The circle described by the singly - ionized charge will have a radius double that of the other circle.

The two circle touch each other because brojected from the same place.

3. Both fields cannot be zero.
4. Both fields can be non zero.

Explanation:

If a charged particle is projected in a gravity free room with some velocity, then its deflection depends on whether the particle experiences electric force or magnetic force. If electric field or magnetic field is absent in the space, then the charged particle will continue to move in the same direction with the same velocity. Thus, both electric and magnetic fields cannot be zero.

If force on the charged particle due to electric field is equal and opposite to the force due to magnetic field, then

$\text{F}=\text{qE}=\text{qvB}$

$\text{v}=\frac{\text{E}}{\text{B}}$

Thus, if anyone of the two fields is non zero, the charged particle will get deflect. Therefore, the correct option is (c) and (d).

1. The electric field must be zero.

4. The magnetic field may or may not be zero.

Explanation:

Force on charged particle in an electric eld, $\text{F} = \text{qE} \ ...(1)$ 

Force on charged particle in a magnetic eld $\text{F} = \text{q} (\text{v}\times\text{b}) = \text{qvB} \sin\theta \ ...(2) $

Where boldface letter represent vector nature of that quantity, q is charge of the particle, v is the velocity of the particle( if any), and $\theta$ is the angle between velocity and magnetic eld. 

From (1), F_E = 0 only when either q = 0 or E = 0. 

Let q ≠ 0, and F ≠ 0, then we must have E ≠ 0 

From (2), if q ≠ 0, v ≠ 0 and B ≠ 0 even then F_B can be 'zero' because of θ = 0^° or 180^°

1. Towards west.

Explanation:

$\text{F}=\text{q}\big(\overrightarrow{\text{V}}\times\overrightarrow{\text{B}}\big)$

$\text{j}=\text{q}\big(\overrightarrow{\text{i}}\times\overrightarrow{\text{B}}\big)$

$\Rightarrow\text{B}\otimes$

⇒ The magnetic field may be down ward direction.

2. $\text{E}=0,\ \text{B}\not=0$

Explanation:

A charged particle moves along a circle that mean Magnetic force is provides centripetal force that causes particle is move in a circle.

So, $\text{E}=0,\ \text{B}\not=0$

3. A helix with uniform pitch.

Explanation:

$\vec{\text{F}}=\text{q}(\vec{\text{V}}\times\vec{\text{B}})=\text{qvB}\sin\theta$

Megnetic force doesn't change the speed of the particle. It change the direction of the velocity of the particle.

V $\cos\theta$ provide the displacement of the particle in Horizontal direction & force is provide the centripetal acceleration of the particle.

So the path of the particle will be a helix with uniform pitch.

1. y-axis.
2. z-axis.

Explanation:

$\text{F}=\text{q}(\overrightarrow{\text{V}}\times\overrightarrow{\text{B}})$

This can be done by applying the Magnetic field along y axis or z axis.

1. $\text{E}=0,\ \text{B}=0$

2. $\text{E}=0,\ \text{B}\not=0$

4.  $\text{E}\not=0,\ \text{B}\not=0$

Explanation:

⇒ Particle move with constant velocity in ay direction. So, B = 0, E = 0

⇒ Particle move in a circle with constant speed. Magnetic force is provide the centripetal force that causes particle is move in a circle.

⇒ If qE = qvB and Magnetic & Electric force in opposite direction in this case particle also move with uniform speed.

4. A helix with nonuniform pitch.

Explanation:

$\text{F}=\text{q}\vec{\text{E}}+\text{q}(\vec{\text{V}}\times\vec{\text{B}})$

$\text{F}=\text{q}\vec{\text{E}}$ provides the acceleration in 'x' direction.

$\text{F}_2=\text{}\text{q}(\vec{\text{V}}\times\vec{\text{B}})$ provides the centripetal Force.

The path of the particle will be ahelix with nonuniform pitch.

4. Li⁺

Explanation:

$\text{T}=\frac{2\pi\text{r}}{\text{v}\bot}\ ...(1)$

$\text{r}=\frac{\text{mv}_1}{\text{qB}}$

$\frac{\text{r}}{\text{v}\bot}=\frac{\text{m}}{\text{qB}}\ ...(2)$

from eq. (1) & (2) we get

$\text{T}=\frac{2\pi\text{m}}{\text{qB}}$

$\text{f}=\frac{1}{\text{T}}=\frac{\text{qB}}{2\pi\text{m}}$

Charge of all these particles are same but mass of Li⁺ is Highest.

 $\therefore\text{ mass}\uparrow,\ \text{f}\downarrow$