Moving Charges and Magnetism

$ \begin{aligned} &\text { Magnetic moment of solenoid }\\ &\begin{aligned} M & =N I A \\ & =500 \times 15 \times \pi \times\left(2 \times 10^{-2}\right)^2 \\ & =3 \pi \mathrm{JT}^{-1} \end{aligned} \end{aligned} $

$ \begin{aligned} B_{\max } & =\frac{\mu_0 I R}{2 \pi R^2}=\frac{\mu_0 I}{2 \pi R} \\ & =\frac{4 \pi \times 10^{-7} \times 12}{2 \times 3.14 \times 0.6 \times 10^{-3}} \\ & =4 \times 10^{-3} \mathrm{~T}=4 \mathrm{mT} \end{aligned} $

Current sensitivity,

$ S=\frac{N B A}{K} $

Where, $K \rightarrow$ Torsional constant

$ \begin{array}{ll} \Rightarrow \quad & S \propto N B A \\ \therefore \quad & \frac{S_A}{S_B}=\left(\frac{N_A}{N_B}\right)\left(\frac{B_A}{B_B}\right)\left(\frac{A_A}{A_B}\right) \\ & =\left(\frac{36}{48}\right)\left(\frac{0.25}{0.5}\right)\left(\frac{2.4 \times 10^{-3}}{4.8 \times 10^{-3}}\right) \\ & =\frac{3}{4} \times \frac{1}{2} \times \frac{1}{2}=\frac{3}{16} \\ \therefore S_A & : S_B=3: 16 \end{array} $

$ \text { } \begin{aligned} & L=\pi r \\ \Rightarrow \quad r & =\frac{L}{\pi} \\ \therefore \quad B & =\frac{\mu_0 I}{4 r}=\frac{\mu_0 I}{4 \times \frac{L}{\pi}}=\frac{\mu_0 \pi I}{4 L} \end{aligned} $

Full scale deflection current,

$ \begin{aligned} I_g & =\frac{30}{0.0625 \operatorname{div} / \mu \mathrm{A}} \\ & =480 \mu \mathrm{~A}=480 \times 10^{-6} \mathrm{~A} \end{aligned} $

Resistance of voltmeter

$ R_V=R_g+R_s $

∴ Maximum measured voltage

$ \begin{aligned} V & =I_g R_V \\ R_V & =\frac{V}{I_g}=\frac{6}{480 \times 10^{-6}} \\ & =12.5 \times 10^3 \Omega=12.5 \mathrm{k} \Omega \end{aligned} $

For the first long solenoid,

$ \begin{aligned} & B_1=\mu_0 n I \\ & 6.24 \times 10^{-2}=\mu_0 \times 400 \times I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & B_2=\mu_0 n_2 i_2 \\ & B_2=\mu_0 200 \times \frac{i}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Dividing (ii) by (i), we get

$ \frac{B_2}{6.24 \times 10^{-2}}=\frac{200}{400} \times \frac{i / 2}{i}=\frac{1}{4} $

$ \begin{aligned}So,\,\, B_2 & =\frac{6.24 \times 10^{-2}}{4} \\ & =1.56 \times 10^{-2} \mathrm{~T} \end{aligned} $

Given:

Kinetic energy of proton

$ K = 8.35\ \text{MeV} $

Magnetic field

$ B = 10\ \text{T} $

Charge of proton

$ q = 1.6 \times 10^{-19}\ \text{C} $

Mass of proton

$ m = 1.67 \times 10^{-27}\ \text{kg} $

The proton enters at right angles to the field.

Step 1: Get proton’s velocity from its kinetic energy

$ K = \frac{1}{2} m v^2 \Rightarrow v = \sqrt{\frac{2K}{m}} $

First, convert $ K $ from MeV to joules:

$ K = 8.35 \times 10^6 \times 1.6 \times 10^{-19} = 1.336 \times 10^{-12}\ \text{J} $

Now compute $ v $:

$ v = \sqrt{\frac{2 \times 1.336 \times 10^{-12}}{1.67 \times 10^{-27}}} $

$ v = \sqrt{1.598 \times 10^{15}} = 3.998 \times 10^{7}\ \text{m/s} $

So $ v \approx 4.0 \times 10^{7}\ \text{m/s} $.

Step 2: Magnetic force on a moving charge

At right angles,

$ F = q v B $

Substituting values:

$ F = (1.6 \times 10^{-19}) (3.998 \times 10^{7}) (10) $

$ F = 6.3968 \times 10^{-11}\ \text{N} $

$ F \approx 6.4 \times 10^{-11}\ \text{N} $

Or equivalently:

$ F = 64 \times 10^{-12}\ \text{N} $

✅ Final Answer:

$ \boxed{F = 64 \times 10^{-12}\ \text{N}} $

Option C is correct.

Radius of circular path in which a charged particle move when it enters a region of perpendicular uniform magnetic field is;

$ \begin{aligned} r & =\frac{m v}{B q}=\frac{\sqrt{2 m K}}{B q} \\ & =\frac{\sqrt{2 m K}}{m \cdot B\left(\frac{q}{m}\right)}=\sqrt{2} \sqrt{\frac{K}{m}} \cdot \frac{1}{B \cdot s} \end{aligned} $

where, $s=\frac{q}{m}=$ specific charge

Ratio of radii of the particles will be

$ \frac{r_1}{r_2}=\sqrt{\frac{m_2}{m_1}\left(\frac{s_2}{s_1}\right)^2} $

( $\therefore K_1=K_2$ and $B$ is same for both particles.)

$ \begin{aligned} \therefore \frac{r_1}{r_2} & =\sqrt{\left(\frac{m_2}{m_1}\right)\left(\frac{S_2}{S_1}\right)^2} \\ & =\sqrt{\left(\frac{4}{1}\right) \cdot\left(\frac{1}{2}\right)^2}=1 \end{aligned} $

An electron moves in a circular orbit of

radius $ R $

and period $ T $.

Step 1: Find the current due to orbital motion

The charge of the electron is $ e $ (magnitude).

In one complete revolution, the charge $ e $ passes a given point once in time $ T $.

So,

$ \text{current } I = \frac{e}{T} $

Step 2: Magnetic moment due to a current loop

Magnetic moment of a current loop is given by

$ \mu = I \times \text{area of the loop} $

The area of a circular loop of radius $ R $ is

$ A = \pi R^2 $

Thus,

$ \mu = I \times A = \frac{e}{T} \times \pi R^2 = \frac{\pi e R^2}{T} $

✅ Correct Option:

Option C → $\displaystyle \frac{\pi e R^2}{T}$

The solenoid has 5 closely packed layers and each layer has 700 turns.

$\therefore$ Total number of turns,

$ \begin{aligned} N & =700 \times 5=3500 \\ \therefore \quad B & =\mu_0 n I \\ & =\mu_0 \times \frac{N}{L} \times I \\ & =4 \pi \times 10^{-7} \times \frac{3500}{1} \times 5 \\ & =21.9 \times 10^{-3} \mathrm{~T} \\ & \simeq 22 \times 10^{-3} \mathrm{~T} \\ & \approx 22 \mathrm{mT} \end{aligned} $

When a charged particle moves into a magnetic field at a right angle (normally), it starts moving in a circle because the magnetic force keeps turning it.

The time it takes to complete one full circle (the time period $T$) is given by:

$T = \frac{2 \pi m}{B q}$

Here, $m$ is the mass of the particle, $q$ is its charge, and $B$ is the magnetic field strength.

You can also write it as:

$T = \frac{2 \pi}{B \cdot \left(\frac{q}{m}\right)}$

This means the time period $T$ is inversely related to the specific charge $\left(\frac{q}{m}\right)$ (charge divided by mass).

So, if the specific charge increases, the time period decreases, and if the specific charge decreases, the time period increases.

Magnetic field at a point inside the wire

$ B_{i n}=\frac{\mu_0 I r}{2 \pi a^2} $

Here, $a \rightarrow$ radius of wire $r \rightarrow$ distance from centre of wire Here, $r=0.5 a$

$ \therefore \quad B_{i n}=\frac{\mu_0 I \cdot(0.5 a)}{2 \pi a^2}=\frac{\mu_0 I}{4 \pi a} $

Magnetic field at a point outside the wire,

$ B_{\text {out }}=\frac{\mu_0 I}{2 \pi r} $

Here, $\quad r=1.5 a$

$ \begin{array}{ll} \therefore & B_{\text {out }}=\frac{\mu_0 I}{2 \pi(1.5 a)}=\frac{\mu_0 I}{3 \pi a} \\ \therefore & \frac{B_{\text {in }}}{B_{\text {out }}}=\frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{3 \pi a}}=\frac{3}{4} \text { or } 3: 4 \end{array} $

$ \text { } \begin{aligned} B & =\frac{\mu_0 I r}{2 \pi R^2} \\ & =\frac{\left(4 \pi \times 10^{-7}\right) \times 2 \times 0.25 \times 10^{-3}}{2 \pi \times\left(1 \times 10^{-3}\right)^2} \\ & =1 \times 10^{-4} \mathrm{~T}=100 \times 10^{-6} \mathrm{~T} \\ & =100 \mu \mathrm{~T} \end{aligned} $

$B_a=\frac{\mu_0 I R^2}{2\left(R^2+a^2\right)^{\frac{3}{2}}}$

Here, $a=R$

$ \begin{aligned} & \therefore \quad B_a=\frac{\mu_0 I R^2}{2\left(R^2+R^2\right)^{\frac{3}{2}}}=\frac{\mu_0 I}{4 \sqrt{2} R} \\ & =\frac{\mu_0 I}{2 R} \cdot \frac{1}{2 \sqrt{2}} \\ & \Rightarrow B_a=\frac{B_c}{2 \sqrt{2}} \quad\left[\because B_c=\frac{\mu_0 I}{2 R}\right] \\ & \Rightarrow \frac{B_c}{B_a}=2 \sqrt{2} \end{aligned} $

Given:

• Current, $ I = 8~\text{A} $

• Magnetic field, $ B = 0.15~\text{T} $

• Angle between wire and field, $ \theta = 30^\circ $

We are asked: force per unit length on the wire.

Formula:

$ \frac{F}{L} = B I \sin \theta $

Substitution:

$ \frac{F}{L} = (0.15)(8)\sin 30^\circ $

$ \sin 30^\circ = \tfrac{1}{2} $

$ \frac{F}{L} = 0.15 \times 8 \times 0.5 $

$ \frac{F}{L} = 0.6~\text{N/m} $

✅ Answer: $0.6~\mathrm{N\,m^{-1}}$

Correct Option: (C)

           $ r=\frac{m v}{B q} $

$ \Rightarrow \quad v=\frac{B q r}{m} $

and  $ \lambda=\frac{h}{m v}=\frac{h}{m \cdot \frac{B q r}{m}} $

                                            [From Eq. (i)]

$ \begin{aligned} & =\frac{h}{B q r} \\ = & \frac{6.63 \times 10^{-34}}{2 \times 10^{-2} \times 2 \times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}} \\ = & 2.07 \times 10^{-10} \mathrm{~m}=2.07 \mathop {\rm{A}}\limits^{\rm{o}}=2.1 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

• Linear mass density of wire, $ \lambda = 0.12~\mathrm{kg/m} $

• Magnetic field, $ B = 0.5~\mathrm{T} $

• Acceleration due to gravity, $ g = 10~\mathrm{m/s^2} $

• The wire is suspended in air → magnetic force per unit length balances weight per unit length.

Step 1: Write force balance

For unit length of wire:

$ \text{Magnetic force per unit length} = \text{Weight per unit length} $

So,

$ I B = \lambda g $

Step 2: Solve for $ I $

$ I = \frac{\lambda g}{B} $

Substitute:

$ I = \frac{0.12 \times 10}{0.5} $

$ I = \frac{1.2}{0.5} = 2.4~\mathrm{A} $

✅ Final Answer: $ \boxed{I = 2.4~\mathrm{A}} $

Option A — 2.4 A

Magnetic field at the centre of the loop $A$,

$ \begin{aligned} B_A & =\frac{\mu_0 I_A}{2 r} \\ & =\frac{4 \pi \times 10^{-7} \times 3}{2 \times 0.02 \times \pi} \\ & =3 \times 10^{-5} \mathrm{~J} \end{aligned} $

The magnetic field at the centre of the loop $B$,

$ \begin{aligned} B_B & =\frac{\mu_0 I_B}{2 r}=\frac{4 \pi \times 10^{-7} \times 4}{2 \times 0.02 \times \pi} \\ & =4 \times 10^{-5} \mathrm{~T} \end{aligned} $

Since, loops $A$ and $B$ are placed at right angles, thus magnetic field $B_A$ and $B_B$ are perpendicular to each other.

Thus, net magnetic field

$ \begin{aligned} B & =\sqrt{B_A^2+B_B^2} \\ & =\sqrt{\left(3 \times 10^{-5}\right)^2+\left(4 \times 10^{-5}\right)^2} \\ & =5 \times 10^{-5} \mathrm{~T} \end{aligned} $

$ \text { } \begin{aligned} B & =\frac{\mu_0 I}{2 \pi r} \\ & =2 \times 10^{-7} \times \frac{4}{10 \times 10^{-2}} \\ & =8 \times 10^{-6} \mathrm{~T}=8 \mu \mathrm{~T} \end{aligned} $

We have a velocity selector in which the electric field E and magnetic field B are perpendicular, and the particle passes undeflected when:

$ qE = qvB $

so

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

Given values

$ v = 6\,\text{km/s} = 6 \times 10^3\,\text{m/s} $

$ E = 400\,\text{V/m} $

Compute $ B $

$ B = \frac{E}{v} = \frac{400}{6 \times 10^3} = \frac{400}{6000} = \frac{4}{60} = \frac{1}{15}\,\text{T} $

✅ Answer:

$ \boxed{B = \frac{1}{15}\, \text{T}} $

Correct Option: (C) $ \frac{1}{15}\, \text{T} $

• Number of turns, $ N = 1200 $

• Area of cross-section, $ A = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2 $

• Magnetic moment, $ M = 1.2 \text{ J T}^{-1} $

We are asked to find the current $ I $.

Step 1: Formula for magnetic moment of a solenoid

A solenoid with $ N $ turns, cross-sectional area $ A $, and current $ I $ has a magnetic moment:

$ M = N I A $

Step 2: Substitute given values

$ 1.2 = 1200 \times I \times (5 \times 10^{-4}) $

Step 3: Simplify

$ 1.2 = 1200 \times 5 \times 10^{-4} \times I $

$ 1.2 = (0.6) \times I $

Step 4: Solve for $ I $

$ I = \frac{1.2}{0.6} = 2 \, \text{A} $

✅ Final Answer:

$ \boxed{I = 2 \, \text{A}} $

Correct option: (B) 2 A

The magnetic dipole moment μ of a rotating charged ring can be found by noting that the rotating charge constitutes a current and then multiplying by the ring’s area:

Current due to rotation

The ring completes one revolution in time

$T = \frac{2\pi}{\omega}\,. $

So the effective current is

$I = \frac{q}{T} = \frac{q\,\omega}{2\pi}\,. $

Area of the ring

$A = \pi R^2\,. $

Magnetic dipole moment

$\mu = I\,A = \frac{q\,\omega}{2\pi}\;\times\;\pi R^2 \;=\;\frac{q\,\omega\,R^2}{2}\,. $

Thus the correct choice is

Option C: $\boxed{\frac{q\,\omega\,R^2}{2}}. $

Given,

Current flows in a conductor from to west.

According to the right-hand rule for a straight conductor, if you point your right thumb in the direction of conventional current, then your curled fingers indicate the direction of the magnetic field.

Here, thumb points in the east-west direction, then according to right-hand rule, the direction of the magnetic field at a point below the conductor is towards south.

We know that, The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by

$ B=\frac{\mu_0}{2 \pi} \cdot \frac{I}{r} $

$ \begin{aligned} &\text { }\\ &\begin{array}{ll} Given, I_1=8 \mathrm{~A} & \text { (along } X \text {-axis) } \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I_2=6 \mathrm{~A} & \text { (along } Y \text {-axis) } \end{array} \end{aligned} $

The direction of magnetic field due to infinite length of wire at point $P$ is shown in figure.

Magnetic field due to wire 1 is,

$ B_1=\frac{\mu_0}{2 \pi} \cdot \frac{8}{d}=\frac{\mu_0}{\pi} \cdot \frac{4}{d} $

Magnetic field due to wire 2 is,

$ B_2=\frac{\mu_0}{2 \pi} \cdot \frac{6}{d}=\frac{\mu_0}{\pi} \cdot \frac{3}{d} $

Net magnetic field at point $P$ is,

$ \begin{aligned} B_{\mathrm{net}} & =\sqrt{B_1^2+B_2^2}=\sqrt{\left(\frac{\mu_0}{\pi} \cdot \frac{4}{d}\right)^2+\left(\frac{\mu_0}{\pi} \cdot \frac{3}{d}\right)^2} \\ & =\frac{\mu_0}{\pi d}\left(\sqrt{4^2+3^2}\right)=\frac{\mu_0}{\pi d} \cdot \sqrt{25} \\ B_{\mathrm{net}} & =\frac{5 \mu_0}{\pi d} \end{aligned} $

Given the details of the problem, we start with the horizontal component of the Earth's magnetic field:

$ B_H = 24 \times 10^{-6} \, \text{T} $

The initial time period of oscillation for the magnet is:

$ T_1 = 0.1 \, \text{s} $

The current flowing through the vertical wire is:

$ I = 18 \, \text{A} $

The distance between the wire and the magnet is:

$ r = 20 \, \text{cm} = 0.2 \, \text{m} $

To find the magnetic field due to the current-carrying wire, we use the formula for the magnetic field around a long straight conductor:

$ B_{\text{wire}} = \frac{\mu_0}{2\pi} \cdot \frac{I}{r} $

Substituting the known values:

$ B_{\text{wire}} = \frac{10^{-7} \times 2 \times 18}{0.2} = 18 \times 10^{-6} \, \text{T} $

The time period of a magnetic dipole oscillating in a magnetic field is given by:

$ T = 2 \pi \sqrt{\frac{I}{MB_H}} $

where:

$ I $ is the moment of inertia,

$ M $ is the magnetic moment.

For the initial and new conditions, we have:

$ T_1 = 2 \pi \sqrt{\frac{I}{M B_H}} \quad \text{(i)} $

$ T_2 = 2 \pi \sqrt{\frac{I}{M (B_H + B_{\text{wire}})}} \quad \text{(ii)} $

By dividing equation (i) by equation (ii), we obtain:

$ \frac{T_1}{T_2} = \sqrt{\frac{B_H + B_{\text{wire}}}{B_H}} $

Substitute the given values:

$ \frac{T_1}{T_2} = \sqrt{\frac{24 \times 10^{-6} + 18 \times 10^{-6}}{24 \times 10^{-6}}} $

$ \frac{0.1}{T_2} = \sqrt{\frac{42 \times 10^{-6}}{24 \times 10^{-6}}} = \frac{\sqrt{7}}{2} $

Solving further:

$ T_2 = \frac{0.2}{\sqrt{7}} $

Thus, the new time period of oscillation of the magnet is approximately:

$ T_2 = 0.076 \, \text{s} $

Number of turns: $N_1 = 400$, $N_2 = 200$

Average radii: $r_1 = 30 \, \text{cm} = 0.3 \, \text{m}$, $r_2 = 60 \, \text{cm} = 0.6 \, \text{m}$

The magnetic field inside a toroid is given by the formula:

$ B = \frac{\mu_0 N I}{2 \pi R} $

where $B$ is the magnetic field, $\mu_0$ is the permeability of free space, $N$ is the number of turns, $I$ is the current, and $R$ is the average radius of the toroid.

For the first toroid, the magnetic field $B_1$ is:

$ B_1 = \frac{\mu_0 N_1 I}{2 \pi R_1} $

For the second toroid, the magnetic field $B_2$ is:

$ B_2 = \frac{\mu_0 N_2 I}{2 \pi R_2} $

To find the ratio $\frac{B_1}{B_2}$:

$ \frac{B_1}{B_2} = \frac{\mu_0 N_1 I}{2 \pi R_1} \times \frac{2 \pi R_2}{\mu_0 N_2 I} = \frac{N_1}{R_1} \times \frac{R_2}{N_2} $

Substituting the values:

$ \frac{B_1}{B_2} = \frac{400}{0.3} \times \frac{0.6}{200} = \frac{4}{1} $

Therefore, the ratio of magnetic fields in the two toroids is $ B_1 : B_2 = 4 : 1 $.

There are three ring having equal radius $r$, they are perpendicular to each other

Current flowing in each ring $=I$ Magnetic field at the common centre is

$ \mathbf{B}=\frac{\mu_0 I}{2 r}( \pm \hat{\mathbf{i}})+\frac{\mu_0 I}{2 r}( \pm \hat{\mathbf{j}})+\frac{\mu_0 I}{2 r}( \pm \hat{\mathbf{k}}) $

Thus, $|\mathbf{B}|=\frac{\mu_0 I}{2 r} \sqrt{3}$ or $|\mathbf{B}|=\frac{\sqrt{3} \mu_0 I}{2 r}$

To determine the ratio of the magnetic forces acting on the proton and the alpha particle, we have the following information:

Ratio of kinetic energy: $(\mathrm{KE})_p : (\mathrm{KE})_\alpha = 1:4$

Magnetic field strength: $B = 3 \, \mathrm{T}$

Angle with the magnetic field: $\theta = 90^\circ$

Definitions:

$q$ is the charge of the particle.

$v$ is the velocity of the particle.

$F$ is the magnetic force acting on the particle.

Using the Lorentz Force Equation:

The magnetic force ($F$) on a charged particle moving in a magnetic field is given by:

$ F = qvB \sin \theta $

Since $\theta = 90^\circ$, $\sin 90^\circ = 1$, so:

$ F = qvB $

Given that the ratio of kinetic energy (KE) is:

$ \frac{(\mathrm{KE})_p}{(\mathrm{KE})_\alpha} = \frac{1}{4} $

The kinetic energy ($\mathrm{KE}$) is defined as:

$ \mathrm{KE} = \frac{1}{2} m v^2 $

Thus, we have:

$ \frac{\frac{1}{2} m_p v_p^2}{\frac{1}{2} m_\alpha v_\alpha^2} = \frac{1}{4} $

Since the mass of the alpha particle $(m_\alpha)$ is four times that of the proton $(m_p)$, i.e., $m_\alpha = 4m_p$, the equation simplifies to:

$ \frac{v_p}{v_\alpha} = 1 \implies v_p = v_\alpha $

Calculating the Ratio of Forces:

From the equation $F = qvB$, the forces on the proton and alpha particle are:

For the proton: $F_p = q_p v_p B$

For the alpha particle: $F_\alpha = q_\alpha v_\alpha B$

The charge of the proton is $q_p = e$ and the charge of the alpha particle is $q_\alpha = 2e$. Therefore, we find the ratio of the forces:

$ \frac{F_p}{F_\alpha} = \frac{q_p v_p B}{q_\alpha v_\alpha B} = \frac{e}{2e} = \frac{1}{2} $

Thus, the ratio of the magnetic force acting on the proton to that on the alpha particle is $1:2$.

Given:

Magnetic field, $ B = 4 \, \text{mT} = 4 \times 10^{-3} \, \text{T} $

Specific charge (charge-to-mass ratio), $ \frac{q}{m} = 8 \times 10^7 \, \text{C/kg} $

When a particle moves in a circular path in an external magnetic field perpendicular to its velocity, the magnetic force acting on it is given by:

$ qvB = \frac{mv^2}{r} $

Also, the relationship between velocity, radius, and angular velocity is:

$ v = r\omega $

Thus, the equation becomes:

$ qvB = m\omega^2r $

Rearranging terms, we find:

$ qvB = m(r\omega)\omega $

$ qvB = mv\omega $

Considering the specific charge, we derive the formula for angular velocity, $ \omega $:

$ \omega = \frac{qB}{m} $

Substituting the given values:

$ \begin{aligned} \omega &= 8 \times 10^7 \times 4 \times 10^{-3} \\ \omega &= 32 \times 10^4 \, \text{rad/s} \end{aligned} $

To find the ratio of the potential differences $\frac{V_2}{V_1}$, we start by considering the kinetic energy relationship for an electron:

$ K = \frac{1}{2} m v^2 = e V $

From this equation, we can solve for $v$:

$ v = \sqrt{\frac{2 e V}{m}} $

The force $F$ exerted by the magnetic field on an electron moving with velocity $v$ is given by:

$ F = e v \times B $

Substituting the expression for $v$ into the force equation:

$ F = e \cdot \sqrt{\frac{2 e V}{m}} \times B $

We are given:

When the potential difference is $V_1$, the force is $F_1 = F$.

When the potential difference is $V_2$, the force doubles, $F_2 = 2F$.

According to the problem, we set up the force ratio:

$ \frac{F_1}{F_2} = \frac{e \cdot \sqrt{\frac{2 e V_1}{m}} \times B}{e \cdot \sqrt{\frac{2 e V_2}{m}} \times B} $

Simplifying, we have:

$ \frac{F}{2F} = \sqrt{\frac{V_1}{V_2}} $

This simplifies to:

$ \frac{1}{2} = \sqrt{\frac{V_1}{V_2}} $

Squaring both sides gives:

$ \frac{V_1}{V_2} = \frac{1}{4} $

This means:

$ \frac{V_2}{V_1} = \frac{4}{1} $

Thus, the ratio $V_2 : V_1$ is $4:1$.

A rectangular loop $ PQRS $ is positioned near a long straight wire as shown.

Dimensions and Parameters:

Length of $ PQ $ and $ RS $: $ l = 25 \, \text{cm} = 0.25 \, \text{m} $

Length of $ QR $ and $ PS $: $ b = 10 \, \text{cm} = 0.10 \, \text{m} $

Distance from $ PQ $ to wire $ XY $: $ r_1 = 10 \, \text{cm} = 0.1 \, \text{m} $

Distance from $ RS $ to wire $ XY $: $ r_2 = 20 \, \text{cm} = 0.2 \, \text{m} $

Calculating Forces:

Force on $ PQ $:

Current in $ PQ $ is antiparallel to the current in $ XY $, causing repulsion. The repulsive force is calculated by:

$ F_1 = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{r_1} \cdot l $

Substituting values:

$ F_1 = \frac{2 \times 10^{-7} \times 25 \times 10 \times 0.25}{0.1} = 1250 \times 10^{-7} = 1.25 \times 10^{-4} \, \text{N} $

Force on $ RS $:

Current in $ RS $ is parallel to the current in $ XY $, causing attraction. The attractive force is:

$ F_2 = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{r_2} \cdot l $

Substituting values:

$ F_2 = \frac{2 \times 10^{-7} \times 10 \times 25 \times 0.25}{0.2} = 625 \times 10^{-7} = 0.625 \times 10^{-4} \, \text{N} $

Forces on $ PS $ and $ QR $:

The forces due to currents in $ PS $ and $ QR $ are equal in magnitude and opposite, cancelling each other out.

Net Force on the Loop:

The net force is the difference between $ F_1 $ and $ F_2 $:

$ F_{\text{net}} = F_1 - F_2 $

$ F_{\text{net}} = 1.25 \times 10^{-4} - 0.625 \times 10^{-4} $

$ F_{\text{net}} = 0.625 \times 10^{-4} = 6.25 \times 10^{-5} \, \text{N} $

The net force is directed away from the long wire $ XY $.

Given, current in conductor $A, i_A=4.5 \mathrm{~A}$

current in conductor $B, i_B=8 \mathrm{~A}$ distance between two conductors,

$ d=25 \mathrm{~cm}=25 \times 10^{-2} \mathrm{~m} $

Let $\odot$ represent the wire. Thus, magnetic field due to a conductor is given by

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

$\therefore$ Magnetic field due to wire $A$

$ B_A=\frac{\mu_0 \times 4.5}{2 \times 314 \times 15 \times 10^{-2}} $

Magnetic field due to wire $B$,

$ \begin{aligned} B_B & =\frac{\mu_0 \times 8}{2 \times 314 \times 20 \times 10^{-2}} \\ B_{\text {net }} & =\sqrt{B_A^2+B_B^2+2 B_A B_A \cos \theta} \end{aligned} $

From $\triangle A P B, \cos \theta=0 \quad \left(\because \theta=90^{\circ}\right)$

$ \begin{aligned} \Rightarrow \quad B_{\mathrm{net}} & =\sqrt{B_A^2+B_B^2} \\ & =\frac{\mu_0}{10^{-2} \times 2 \pi} \sqrt{\left(\frac{4.5}{15}\right)^2+\left(\frac{8}{20}\right)^2} \\ & =\frac{\mu_0}{2 \pi \times 10^{-2}} \times \sqrt{(0.3)^2+(0.4)^2} \\ & =\frac{\mu_0}{2 \pi \times 10^{-2}} \times(0.5) \\ & =1 \times 10^{-5} \mathrm{~N} \end{aligned} $

Given:

Radius of the first ring, $ R_1 = 40 \, \text{cm} = 0.4 \, \text{m} $

Radius of the second ring, $ R_2 = 50 \, \text{cm} = 0.5 \, \text{m} $

Current carried by each ring, $ I = 3.5 \, \text{A} $

The magnetic field at the center of a circular coil is given by:

$ B_1 = \frac{\mu_0 I}{2 R_1} $

For the second coil:

$ B_2 = \frac{\mu_0 I}{2 R_2} $

Since the currents are in opposite directions, the net magnetic field at the center is:

$ \begin{align*} B_{\text{net}} &= B_1 - B_2 = \frac{\mu_0 I}{2} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right] \\ &= \frac{\mu_0 I}{2} \left[ \frac{1}{0.4} - \frac{1}{0.5} \right] \\ &= \frac{4 \pi \times 10^{-7} \times 3.5}{2} \left[ \frac{1}{0.4} - \frac{1}{0.5} \right] \\ &= 1.1 \times 10^{-6} \, \text{T} \\ &= 11 \times 10^{-7} \, \text{T} \end{align*} $

To find the magnetic field induction produced at the center of an electron orbiting a hydrogen nucleus, we start by considering the electron as equivalent to a current loop.

The current ($I$) generated by the electron can be calculated using the formula:

$ I = \frac{q}{t} = q \cdot f $

where:

$ q = 1.6 \times 10^{-19} \, \mathrm{C} $ (charge of an electron),

$ f = 6.6 \times 10^{15} \, \mathrm{Hz} $ (frequency, since the electron completes $6.6 \times 10^{15}$ revolutions per second).

Calculating the current:

$ I = 1.6 \times 10^{-19} \, \mathrm{C} \times 6.6 \times 10^{15} \, \mathrm{Hz} = 1.056 \times 10^{-3} \, \mathrm{A} \approx 1 \, \mathrm{mA} $

The electron orbits a nucleus with a radius of $0.47 \, \mathring{A} = 0.47 \times 10^{-10} \, \mathrm{m}$.

The magnetic field induction ($B$) at the center of the orbit is given by:

$ B = \frac{\mu_0 I}{2r} $

where:

$ \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T} \cdot \mathrm{m/A} $ (permeability of free space),

$ r = 0.47 \times 10^{-10} \, \mathrm{m} $.

Plugging in the values:

$ B = \frac{4\pi \times 10^{-7} \times 10^{-3}}{2 \times 0.47 \times 10^{-10}} $

This simplifies to:

$ B = \frac{4\pi}{2 \times 0.47} \times 10^7 $

Calculating further:

$ B = 1336 \, \mathrm{Wb/m}^2 \approx 14 \, \mathrm{Wb/m}^2 $

Therefore, the magnetic field induction at the center of the electron's orbit is approximately $14 \, \mathrm{Wb/m}^2$.

When two coils are brought near to each other, then magnetic flux linked with both the coils changes, due to which an induce current produces. So, the current in both the coils decreases because induce current opposes the main current.

The given situation is shown below.

$ \begin{aligned} \frac{O M}{A M} & =\tan 60^{\circ} \\ O M & =A M \tan 60^{\circ} \\ & =1 \times \sqrt{3} \mathrm{~cm}=\sqrt{3} \mathrm{~cm} \\ O M & =\sqrt{3} \times 10^{-2} \mathrm{~m} \end{aligned} $

$\therefore$ Magnetic field at the centre,

$ \begin{aligned} B & =6 \times \frac{\mu_0}{4 \pi} \frac{I}{(O M)}\left(\sin 30^{\circ}+\sin 30^{\circ}\right) \\ & =6 \times 10^{-7} \times \frac{4}{\sqrt{3} \times 10^{-2}}\left(\frac{1}{2}+\frac{1}{2}\right) \\ & =8 \sqrt{3} \times 10^{-5} \mathrm{~T} \end{aligned} $

To calculate the magnetic field at the center of a tightly wound coil, we use the formula:

$ B = \frac{\mu_0 N I}{2 r} $

$ \mu_0 $ is the permeability of free space, approximately $ 4\pi \times 10^{-7} \, \text{T m/A} $.

$ N $ is the number of turns in the coil, which is 200.

$ I $ is the current passing through the coil, which is 5 A.

$ r $ is the radius of the coil in meters, given as 20 cm or 0.2 m.

Substituting these values into the formula:

$ B = \frac{4\pi \times 10^{-7} \times 200 \times 5}{2 \times 0.2} $

Solving this, we get:

$ B = 3.14 \times 10^{-3} \, \text{T} $

Thus, the magnetic field at the center of the coil is $ 3.14 \times 10^{-3} \, \text{T} $.

A coil carrying a current experiences a torque when placed in a magnetic field. This torque is quantified by the following relation:

$ \tau = N \cdot I \cdot A \cdot B \cdot \sin \theta $

where $ N $ is the number of turns in the coil, $ I $ is the current flowing through the coil, $ A $ is the area of the coil, $ B $ is the magnetic field strength, and $ \theta $ is the angle between the magnetic field and the normal to the plane of the coil.

The torque is at its maximum when $\sin \theta = 1$, which occurs at $\theta = 90^{\circ}$. Thus, the maximum torque $\tau_{\max}$ can be expressed as:

$ \tau_{\max} = N \cdot I \cdot A \cdot B \quad \ldots \text{(i)} $

According to the problem, the experienced torque is 80% of this maximum:

$ N \cdot I \cdot A \cdot B \cdot \sin \theta = 0.8 \cdot \tau_{\max} $

Substituting $\tau_{\max}$ from equation (i):

$ N \cdot I \cdot A \cdot B \cdot \sin \theta = \frac{80}{100} \cdot (N \cdot I \cdot A \cdot B) $

This simplifies to:

$ \sin \theta = \frac{4}{5} $

We know that:

$ \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{3}{5} $

Thus, the tangent of the angle $\theta$ becomes:

$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{4/5}{3/5} = \frac{4}{3} $

Therefore, the angle $\theta$ is:

$ \theta = \tan^{-1} \left(\frac{4}{3}\right) $

Given:

Velocity of the electron: $ \mathbf{v} = (2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}) \text{ m/s} $

Electric field: $ \mathbf{E} = (3 \hat{\mathbf{i}} + 6 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) \text{ V/m} $

Magnetic field: $ \mathbf{B} = (2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) \text{ T} $

The Lorentz force is calculated using the formula:

$ \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) $

First, solve for $\mathbf{v} \times \mathbf{B}$:

$ \begin{aligned} \mathbf{v} \times \mathbf{B} &= \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 0 \\ 0 & 2 & 3 \end{array}\right| \\ &= (9 \hat{\mathbf{i}} - 6 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}}) \end{aligned} $

Now, find the total force $\mathbf{F}$:

$ \begin{aligned} \mathbf{F} &= q\left[(3 \hat{\mathbf{i}} + 6 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) + (9 \hat{\mathbf{i}} - 6 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}})\right] \\ &= q[12 \hat{\mathbf{i}} + 6 \hat{\mathbf{k}}] \end{aligned} $

For an electron, $ q = 1.6 \times 10^{-19} \text{ C} $:

$ \begin{aligned} \mathbf{F} &= 1.6 \times 10^{-19} \times [12 \hat{\mathbf{i}} + 6 \hat{\mathbf{k}}] \\ |\mathbf{F}| &= 1.6 \times 10^{-19} \times \sqrt{12^2 + 6^2} \\ |\mathbf{F}| &= 1.6 \times 10^{-19} \times \sqrt{180} \\ |\mathbf{F}| &= 2.15 \times 10^{-18} \text{ N} \end{aligned} $

To determine the direction of $\mathbf{F}$ with respect to the $X$-axis:

$ \theta = \tan^{-1}\left(\frac{6}{12}\right) = \tan^{-1}\left(\frac{1}{2}\right) $

Thus, we have:

$ \cos \theta = \frac{2}{\sqrt{5}} \quad \Rightarrow \quad \theta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) $

Kinetic energy of electron,

$\begin{aligned} K & =100 \mathrm{eV} \\ & =100 \times 1.6 \times 10^{-19} \mathrm{~J}=1.6 \times 10^{-17} \mathrm{~J} \end{aligned}$

Radius of circular path, $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Mass of electron, $m=0.5 \mathrm{~MeVc}^{-2}$

$\begin{aligned} & =\frac{0.5 \mathrm{MeV}}{c^2}=\frac{0.5 \times 10^6 \times 1.6 \times 10^{-19}}{3 \times 10^8 \times 3 \times 10^8} \mathrm{~kg} \\ & =8.89 \times 10^{-31} \mathrm{~kg} \end{aligned}$

$\because$ Radius of circular path of electron in magnetic field $B$ in terms of kinetic energy $(K)$ is given as

$\begin{aligned} r =\frac{\sqrt{2 m K}}{B q} \\ \Rightarrow B =\frac{\sqrt{2 m K}}{r q}=\frac{\sqrt{2 \times 8.89 \times 10^{-31} \times 1.6 \times 10^{-17}}}{0.1 \times 1.6 \times 10^{-19}} \\ =3.3 \times 10^{-4} \mathrm{~T} \end{aligned}$

Given :

Mass of the particle, $m = 2.2 \times 10^{-30} \, \mathrm{kg}$

Charge, $q = 1.6 \times 10^{-19} \, \mathrm{C}$

Velocity, $v = 10 \, \text{km/s} = 10^4 \, \mathrm{m/s}$

Radius of the circular path, $r = 2.8 \, \mathrm{cm} = 2.8 \times 10^{-2} \, \mathrm{m}$

Turn density, $n = 25 \, \text{turns/cm} = 2500 \, \text{turns/m}$

Permeability of free space, $\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{H/m}$

The radius of the circular path is given by the equation :

$r = \frac{m v}{B q} \quad \text{(i)}$

The magnetic field inside the solenoid is given by :

$B = \mu_0 n I \quad \text{(ii)}$

Combining equations (i) and (ii), we get :

$\begin{aligned} r &= \frac{m v}{(\mu_0 n I) q} \\\\ \Rightarrow I &= \frac{m v}{\mu_0 n q r} \\\\ &= \frac{2.2 \times 10^{-30} \times 10^4}{4 \pi \times 10^{-7} \times 2500 \times 1.6 \times 10^{-19} \times 2.8 \times 10^{-2}} \\\\ &= 1.56 \times 10^{-3} \, \mathrm{A} = 1.56 \, \mathrm{mA} \end{aligned}$

For a toroid,

Inner radius, $r_1=24 \mathrm{~cm}$

Outer radius, $r_2=25 \mathrm{~cm}$

Current, $I=12 \mathrm{~A}$

$\because$ Length of toroid,

$\begin{aligned} l & =2 \pi \frac{\left(r_1+r_2\right)}{2} \\ & =\pi\left(r_1+r_2\right) \\ & =\pi(24+25) \\ & =0.49 \pi \mathrm{m} \end{aligned}$

$\because$ no. of turns per unit length in toroid,

$\begin{aligned} n & =\frac{N}{l} \\ & =\frac{4900}{0.49 \pi}=\frac{10^4}{\pi} \end{aligned}$

$\because$ Magnetic field inside the core of toroid

$\begin{aligned} B & =\mu_0 n I \\ & =4 \pi \times 10^{-7} \times \frac{10^4}{\pi} \times 12 \\ & =48 \times 10^{-3} \mathrm{~T}=48 \mathrm{mT} \end{aligned}$

The given situation is shown below

Magnetic field intensity at point $O$ due to wire 1

$\mathbf{B}_0=\frac{\mu_0}{2 \pi} \times \frac{\mathbf{I}}{r_1} \hat{k}$

$\Rightarrow \quad \mathbf{B}_0=\frac{\mu_0}{2 \pi} \times \frac{\mathbf{I}}{1 \times 10^{-2}} \hat{k}\quad \text{... (i)}$

Similarly, magnetic field at point $O$ due wire 2

$\begin{aligned} & \quad \mathbf{B}^{\prime}=\frac{\mu_0}{2 \pi} \hat{k} \frac{I}{r_2}=\frac{\mu_0}{2 \pi} \cdot \frac{I}{2 \times 10^{-2}}(-\hat{k}) \\ & \Rightarrow \quad \\ & \mathbf{B}^{\prime}=\frac{1}{2}\left[\frac{\mu_0}{2 \pi} \frac{I}{10^{-2}}\right](\hat{k}) \\ &=\frac{1}{2} \times B_0(-\hat{k}) \quad \text{[Form Eq. (i)]}\\ &=\frac{B_0}{2}(-\hat{k})=-\frac{B_0}{2} \end{aligned}$

Net magnetic field at point $O$.

$\begin{aligned} \mathbf{B}_{\mathrm{net}} & =\mathbf{B}_0+\mathbf{B}^{\prime}=\mathbf{B}_0+\mathbf{B}^{\prime} \\ & =B_0-\frac{B_0}{2}=\frac{B_0}{2} \\ \mathbf{B}_{\mathrm{net}} & =\frac{1}{2} \cdot \frac{\mu_0}{2 \pi} \cdot \frac{1}{1 \times 10^{-2}} \hat{k} \quad \text{[From Eq. (i)]}\\ & =\frac{\mathbf{B}_0}{2} \hat{k} \end{aligned}$

force on the electron,

$\begin{aligned} \mathbf{F} & =-e\left(\mathbf{u} \times \mathbf{B}_{\mathrm{net}}\right) \\ & =-e\left[\mathbf{u}\left(\cos 45 \hat{i}+\sin 45^{\circ} \hat{j}\right) \times \mathbf{B}_{\mathrm{net}}\right] \\ & =-e\left[\frac{\mathbf{u}}{\sqrt{2}}(\hat{i}+\hat{j}) \times \frac{B_0}{2} \hat{k}\right]=\frac{-e u B_0}{2 \sqrt{2}}[\hat{i} \times \hat{k}+\hat{j} \times \hat{k}] \\ & =\frac{-e u B_0}{2 \sqrt{2}}[-\hat{j}+\hat{i}]=-\frac{-e u B_0}{2 \sqrt{2}}(\hat{i}-\hat{j}) \end{aligned}$

If $v_1$ and $v_2$ be the velocities of electron $e_1$ and $e_2$. Then according to given condition.

$\begin{aligned} K_1 & =2 K_2 \\ \frac{1}{2} m v_1^2 & =2\left(\frac{1}{2} m v_2^2\right) \\ \Rightarrow \quad v_1^2 & =2 v_2^2 \Rightarrow v_1=\sqrt{2} v_2 \end{aligned}$

Since, frequency of rotation of electrons in magnetic field is given as,

$f=\frac{2 \pi m}{B q} \Rightarrow f \propto \frac{m}{q}$

Since $m$ and $q$ are same for both electrons $e_1$ and $e_2$ an $f$ does not depend on the velocity of moving electron.

Hence, $f_1=f_2$

Given, number of turns for coil, $N_1=10$

Coil area, $A_1=2 \times 10^{-4} \mathrm{~m}^2$

Coil current, $I_1=0.5 \mathrm{~A}$

and number of turns per metre of solenoid, $n_2=1000$

Current through solenoid, $I_2=3 \mathrm{~A}$

Since, Torque $(\tau)=$ magnetic moment $(M) \times$ magnetic field

$\begin{aligned} \Rightarrow \tau & =I_1 N_1 A_1 \times n_2 \mu_0 I_2 \\ \Rightarrow \tau & =\frac{5}{10} \times 10 \times 2 \times 10^{-4} \times 10^3 \times 4 \pi \times 10^{-7} \times 3 \\ & =12 \pi \times 10^{-7} \mathrm{~N}-\mathrm{m} \end{aligned}$

Given, radius of coil, $R=4 \pi \times 10^{-2} \mathrm{~m}$

Current through coils $I_1=10 \mathrm{~A}, I_2=24 \mathrm{~A}$

Let, $B_1, B_2$ is the magnetic induction due to coil 1 and 2

$B_1=\frac{1}{2} \frac{\mu_0 I_1}{R} \Rightarrow B_1=\frac{1}{2} \times \frac{4 \pi \times 10^{-7} \times 10}{4 \pi \times 10^{-2}}=5 \times 10^{-5} \mathrm{~T} $

Similarly,

$\begin{aligned} B_2 & =\frac{1}{2} \frac{4 \pi \times 10^{-7} \times 24}{4 \pi \times 10^{-2}}=12 \times 10^{-5} \mathrm{~T} \\ \therefore \quad B_{\text {net }} & =\sqrt{B_1^2+B_2^2}=10^{-5} \sqrt{5^2+12^2} \\ & =10^{-5} \sqrt{25+144}=13 \times 10^{-5} \mathrm{~T} \end{aligned}$

Given,

Length of wire $=L$

Current through wire $=I$

$\because$ Magnetic moment, $m=$ Current $(I) \times \operatorname{Area}(A)$ ..... (i)

and since circle of radius $r$ is made of wire of length $L$

$\therefore \quad 2 \pi r=L \Rightarrow r=\frac{L}{2 \pi}$

So, area, $A=\pi r^2$

$=\pi\left(\frac{L}{2 \pi}\right)^2=\frac{L^2}{4 \pi} $

Put this value in Eq. (i), we get

$m=\frac{I L^2}{4 \pi}$

According to given diagram,

As we know that, $F=\frac{\propto_0}{2 \pi} \cdot \frac{I_1 I_2}{d} l$

where, $F$ is force, $\propto_0$ is $4 \pi \times 10^{-7}, I_1$ and $I_2$ are currents in conductors, $l$ is length of conductor and $d$ is separation between two wires. As same sense of current, attract whereas opposite sense, repel.

$\begin{aligned} & \therefore \text { Net force, } F_{\text {net }}=\left(\frac{\alpha_0}{2 \pi} \frac{I_1 I_2}{d_1}-\frac{\alpha_0}{2 \pi} \frac{I_1 \cdot I_2}{d_2}\right) l \\\\ & \quad=\frac{\alpha_0}{2 \pi} \cdot\left(\frac{2 \times 1}{2 \times 10^{-2}}-\frac{2 \times 1}{12 \times 10^{-2}}\right) 15 \times 10^{-2} \\\\ & \quad=25 \times 10^{-7} \mathrm{~N} \text {, towards wire }\end{aligned}$

According to question, construction of current and conductor is

$\because$ Magnetic field due to straight conductor,

$(B)=\frac{1}{2\pi}\frac{\mu_0I}{R}$

Let $B_{net}$ , B$_1$ and B$_2$ are net magnetic field, magnetic field due to conductor 1 and 2, respectively.

$\therefore$ $B_{\text {net }}=B_1-B_2=\frac{\mu_0 I}{2 \pi}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

$\because$ $r_2 > r_1$

$\therefore B_{net}$ will only zero outside conductor or cable.

Given, initial magnetic field at centre of single turn coil $=B$

Final number of turn, $n_2=2$

Let final magnetic field be $B^{\prime}$

$\because \quad B=\frac{\mu_0}{2} \frac{I}{R}$

where, $\mu_0$ is permeability,

$I$ is current,

$R$ is radius,

But length $l$ of conductor is same.

$\Rightarrow \quad l=2 \pi R=2 \times 2 \pi R^{\prime} \Rightarrow R^{\prime}=R / 2$

$\begin{aligned} & \text { Now, } \quad B \propto \frac{n}{R} \\ & \therefore \quad \frac{B^{\prime}}{B}=\frac{2}{\frac{R}{2}} \times \frac{R}{1}=4 \Rightarrow B^{\prime}=4 B\end{aligned}$