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.

Solution:

Magnetic force on a conductor of length l carrying a current I placed in a uniform magnetic field B is given by, $\text{F}=\text{I}(\text{l}\times\text{B})\text{ or }|\text{F}|=\text{I }|\text{l}| \ \ |\text{B}|\sin\theta$.

The direction of force is given by Fleming's left hand rule & F is perpendicular to the direction of magnetic field B, Therefore, work done by the magnetic force on the ions is zero.

3. 2

Explanation:

The magnetic field due to the current-carrying long, straight wire at point a is given by,

$\text{B}=\frac{\mu_0\text{i}}{2\pi\text{d}}$

When both the wires carry currents i₁ and i_2 ​in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.

$\text{B}'=\frac{\mu_0\text{i}_1}{2\pi\text{a}}-\frac{\mu_0\text{i}_2}{2\pi\text{a}}=10\mu\text{T}\ ...(1)$

Here, a is the distance of the midpoint from both the wires.

When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.

$\text{B}''=\frac{\mu_0\text{i}_1}{2\pi\text{a}}+\frac{\mu_0\text{i}_2}{2\pi\text{a}}=30\mu\text{T}\ ...(2)$

On solving eqs. (1) and (2), we get

$\text{i}_1-\text{i}_2=10$

$\text{i}_1+\text{i}_2=30$

$\Rightarrow\text{i}_1=20$

$ \text{i}_2=10$

$\frac{\text{i}_1}{\text{i}_2}=\frac{20}{10}$

$\frac{\text{i}_1}{\text{1}_2}=\frac{2}{1}$

$\Rightarrow\frac{\text{i}_1}{\text{i}_2}=2$

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

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

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

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

2. Minimum at the axis of the wire.
3. Maximum at the surface of the wire.

Explanation:

A long, straight wire of radius R is carrying current i, which is uniformly distributed over its cross section. According to Ampere's law,

$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}_\text{inside}$

At surface,

$\text{B}\times2\pi\text{R}=\mu_0\text{i}$

$\Rightarrow\text{B}_\text{surface}=\frac{\mu_0\text{i}}{2\pi\text{R}}$

Inside, $\text{B}\times2\pi\text{r}=\mu_0\text{i}\ \text{for}\ \text{r}<\text{R}$

Here i^, is the current enclosed by the amperian loop drawn inside the wire.

B_inside will be proportional to the distance from the axis.

On the axis

B = 0

The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre.

Outside, $\text{B}\times2\pi\text{r}=\mu_0\text{i}$

$\Rightarrow\text{B}_\text{outside}=\frac{\mu_0\text{i}}{2\pi\text{r}},\ \text{r}>\text{R}$

Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.

2. $\text{ni}$

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

Explanation:

Lorentz Force:

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

⇒ Dimensions of $\text{x}=[\text{v}]=\Big[\frac{\text{E}}{\text{B}}\Big]=\big[\text{LT}^{-1}\big]$

$\text{y}=\frac{1}{\sqrt{\mu_0\in_0}}=\sqrt{\frac{4\pi}{\mu_0}\times\frac{1}{4\pi\in_0}}=\sqrt{\frac{9\times10^9}{10^{-7}}}=3\times10^8=\text{c}$

⇒ Dimensions of $\text{y}=[\text{c}]=\big[\text{LT}^{-1}\big]$

Time constant of RC circuit = RC so dimensionally [RC] = [T]

$\Rightarrow\text{z}=\Big[\frac{\text{l}}{\text{RC}}\Rightarrow[\text{z}]=[\text{LT}^{-1}]$

Therefore, x, y and z 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 $\overrightarrow{\text{B}}$

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 $\overrightarrow{\text{E}}.$

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

Explanation:

According to Ampere's law, in a coaxial, straight cable carrying currents i in the inner conductor and -i (equally in the opposite direction) in the outside conductor.

Inside the inner conductor

$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}_\text{inside}$

$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=0$

$\Rightarrow\text{b.l}=0$

$\Rightarrow\text{B}=0$

In between the 2 conductors

$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}$

$\Rightarrow\text{B}=\frac{\mu_0\text{i}}{2\pi\text{r}}$

Outside the outer conductor

$\oint\overrightarrow{\text{B}}.\text{d}\overrightarrow{\text{l}}=\mu_0\text{i}$

$\Rightarrow\text{B}=0$

Therefore, the magnetic field is zero outside the cable.

3. $\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$

Explanation:

Particles X and Y of respective masses m₁ and m₂ are carrying charge q describing circular paths with respective radii R₁ and R₂ such that,

$\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}$

$\text{R}_1=\frac{\text{m}_2\text{v}_2}{\text{qB}}$

Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.

$\therefore\frac{1}{2}\text{m}_1\text{v}_1^2=\frac{1}{2}\text{m}_2\text{v}_2^2$

$\because\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}\Rightarrow\text{v}_1=\frac{\text{R}_1\text{qB}}{\text{m}_1}$

And,

$\text{R}_2=\frac{\text{m}_2\text{v}_2}{\text{qB}}\Rightarrow\text{v}_2=\frac{\text{R}_2\text{qB}}{\text{m}_2}$

$\therefore\text{m}_1\Big(\frac{\text{R}_1\text{qB}}{\text{m}_1}\Big)^2=\text{m}_2\Big(\frac{\text{R}_2\text{qB}}{\text{m}_2}\Big)^2$

$\Rightarrow\frac{\text{m}_1}{\text{m}_2}=\frac{\text{R}_1^2}{\text{R}_2^2}=\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$

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

Solution:

The gyro-magnetic ratio of an electron in an H-atom is equal to the ratio of the magnetic moment and the angular momentum of the electron. Both are related as,

$\mu_\text{l}=\Big(\frac{\text{e}}{2\text{m}_\text{e}}\Big)\text{l}$

The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.

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)