Physics M.C.Q (1 Marks)

2. Some charges inside the wire move to the surface as a result of B.​​​​​​

4. If the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire.

1. Towards west.

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

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

2. The magnetic forces on both the particles cause equal accelerations.
3. Both particles gain or loose energy at the same rate.
4. The motion of the centre of mass (CM) is determined by B alone.

4. The charge to mass ratio satisfy: $\Big(\frac{\text{e}}{\text{m}}\Big)_1+\Big(\frac{\text{e}}{\text{m}}\Big)_2=0$.

2. $\text{ni}$

1. x, y have the same dimensions.
2. y, z have the same dimensions.
3. z, x have the same dimensions.

1. $\overrightarrow{\text{E}}||\overrightarrow{\text{B}},\ \overrightarrow{\text{v}}||\overrightarrow{\text{E}}$
2. $\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}}$

1. Outside the cable.
2. Inside the inner conductor.

4. $\text{Zero}$

1. Independent of which orbit it is in.
2. Negative.

1. $\frac{\mu_0\text{i}}{4\pi}\frac{\text{d}\overrightarrow{\text{l}}\times\overrightarrow{\text{r}}}{\text{r}^3}$
2. $-\frac{\mu_0\text{i}}{4\pi}\frac{\overrightarrow{\text{r}}\times​​\text{d}\overrightarrow{\text{l}}}{\text{r}^3}$

3. A helix with uniform pitch.

1. F = 0

2. Is constant inside the tube.
3. Is zero at the axis.

1. Equal and opposite

2. $\text{i}\propto\theta$

Key concept: This problem is based upon the single moving charge placed with some uniform electric and magnetic fields in space. Then they experiences a force called Lorentz force is given by the relation $F_{net} = qE + q(v \times B).$
Force experience by the charged particle due to electric field $F_e = qE$
Force experience by the charged particle due to magnetic field, $F_m = q(v \times B)$
According to the problem, particle is moving with constant velocity means acceleration of particle is zero and also it is not changing its direction of motion.
This will happen when net force on particle is zero.

1. If $E = 0,$ and $v \| B,$ then$ F_{net} = 0.$
2. If $E \neq 0, B \neq 0$ and $E, v$ and $B$ are mutually perpendicular.
3. When both $E$ and $B$ are absent.

2. $\text{B}\perp\text{V}.$

4. Zero.

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

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.

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

1. $\frac{\mu_0\text{i}}{2\pi\text{r}}$

1. Very nearly $2\pi\text{aiB}$ perpendicular to the plane of the wire.

2. 1 : 1 : 2

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 $q \neq 0$, and $F \neq 0,$ then we must have $E \neq 0$
From $(2),$ if $q \neq 0, v \neq 0$ and $B \neq 0$ even then $F_B$ can be 'zero' because of $\theta = 0^\circ$ or $180^\circ $​​​​​​​

3. Will not exert any force on the circular loop.

1. B ⊥ v.

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

4. Zero.

2. $\cos\theta$

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

4.  $\text{E}\not=0,\ \text{B}\not=0$

4. A helix with nonuniform pitch.

2. Towards 40A.

3. Remain unchanged.

2. A moving-coil galvanometer.

3. The kinetic energy.