Moving Charges and Magnetism

We have to find force on length $A C$

$ \begin{aligned} \therefore \quad F & =2 \times 2 \times \frac{30 \sqrt{3}}{3} \times 10^{-2} \times \frac{\sqrt{3}}{2} \\ & =60 \times 10^{-2}=0.6 \mathrm{~N} \end{aligned} $

As, magnetic force is providing centripital acceleration,

$ \begin{aligned} F & =\frac{m v^2}{r} \\ \text { or } B q v \sin \theta & =\frac{m v^2}{r}\left(\theta=90^{\circ}\right) \\ \Rightarrow \quad B & =\frac{m v}{q r} \end{aligned} $

Substituting values,

$ \begin{aligned} B & =\frac{1.66 \times 10^{-27} \times 8 \times 10^5}{1.6 \times 10^{-19} \times 8.3 \times 10^{-2}} \\ & =1 \times 10^{-1} \mathrm{~T}=100 \mathrm{mT} \end{aligned} $

Magnetic moment, $\mu=n I \times A$

For

$ \begin{aligned} \frac{\mu_1}{\mu_2} & =\frac{n_1 I_1 A_1}{n_2 I_2 A_2}=\frac{300}{200} \times \frac{I}{I} \times \frac{\pi r_1^2}{\pi r_2^2} \\ \frac{1}{2} & =\frac{3}{2} \times\left(\frac{r_1}{r_2}\right)^2 \Rightarrow \frac{r_1}{r_2}=\frac{1}{\sqrt{3}} \end{aligned} $

Step 1: Find the magnetic field due to the first wire (8 A)

The formula for the magnetic field at a distance $ r $ from a straight wire carrying current $ I $ is:
$ B = \frac{\mu_0 I}{2\pi r} $
For the first wire: $ I_1 = 8\,A $, $ r_1 = 4\,cm = 4 \times 10^{-2}\,m $
So,
$ \mathbf{B}_1 = \frac{\mu_0 \times 8}{2\pi \times 4 \times 10^{-2}}(-\hat{\mathbf{k}}) $ (The $-\hat{\mathbf{k}}$ shows the direction of the field.)

Step 2: Find the magnetic field due to the second wire (10 A)

The second wire has $ I_2 = 10\,A $ and is $ r_2 = 5\,cm = 5 \times 10^{-2}\,m $ away from the point. (Since the wires are 9 cm apart and the point is 4 cm from the first wire, it must be 5 cm from the second wire.)

So, $ \mathbf{B}_{\mathrm{II}} = \frac{\mu_0 \times 10}{2\pi \times 5 \times 10^{-2}}(-\hat{\mathbf{k}}) $

Step 3: Add the two magnetic fields

Both fields point in the same direction ($-\hat{\mathbf{k}}$), so add their values:

$ \mathbf{B}_{\mathrm{net}} = \mathbf{B}_1 + \mathbf{B}_{\mathrm{II}} \\ = \frac{\mu_0}{2\pi \times 10^{-2}} \left[ \frac{8}{4} + \frac{10}{5} \right] (-\hat{\mathbf{k}}) $

Calculate inside the brackets:

$ \frac{8}{4} = 2,\ \frac{10}{5} = 2 $

So, $ \mathbf{B}_{\mathrm{net}} = \frac{\mu_0}{2\pi \times 10^{-2}} [2+2] (-\hat{\mathbf{k}}) = \frac{\mu_0}{2\pi \times 10^{-2}} \cdot 4 (-\hat{\mathbf{k}}) $

Step 4: Substitute the value of $ \mu_0 $

We know $ \mu_0 = 4\pi \times 10^{-7} $ Tm/A.
So, $ \mathbf{B}_{\mathrm{net}} = \frac{4\pi \times 10^{-7}}{2\pi \times 10^{-2}} \cdot 4 (-\hat{\mathbf{k}}) $

The $ \pi $ cancels: $ \frac{4}{2} = 2 $

So, $ = \frac{2 \times 10^{-7} \times 4}{10^{-2}} (-\hat{\mathbf{k}}) = \frac{8 \times 10^{-7}}{10^{-2}} (-\hat{\mathbf{k}}) = 8 \times 10^{-5}~\text{T} (-\hat{\mathbf{k}}) $

A magnetic Lorentz force will act on the $\alpha$-particle in magnetic field, which provide necessary centripetal force to move an uniform circular path in magnetic field. According to Flemming's left hand rule, $\alpha$-particle will move on horizontal circular path with the same speed.

Magnetic intensity $H$ does not depend on the type of material inside the solenoid.

It only depends on:

• The number of turns per metre ($n$)
• The current passing through the solenoid ($I$)

The formula for magnetic intensity is:

$ H = nI $

Now, substitute the given values:

• $n = 1000$ turns per metre
• $I = 2$ A

So,

$ \begin{aligned} H &= 1000 \times 2 \\ &= 2000~\mathrm{A}/\mathrm{m} \\ &= 2 \times 10^3~\mathrm{Am}^{-1} \end{aligned} $

$ \text { } \begin{aligned} & B=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}} \\ & \text { } \begin{aligned}here,\,\, x & =\sqrt{2} d=\sqrt{2}(2 R) \,\,\,\,\,\,\,\,\,(\because d=2 R)\\ & =2 \sqrt{2} R \\ \therefore \quad B & =\frac{\mu_0 I R^2}{\left.2 R^2+(2 \sqrt{2})^2\right]^{3 / 2}} \\ & =\frac{\mu_0 I R^2}{2\left[R^2+8 R^2\right]^{3 / 2}} \\ & =\frac{\mu_0 I R^2}{2\left[(3 R)^2\right]^{3 / 2}} \\ & =\frac{\mu_0 I R^2}{2 \times(3 R)^3} \\ B & =\frac{\mu_0 I}{54 R}=\frac{1}{27} \times\left(\frac{\mu_0 I}{2 R}\right) \\ B & =\frac{1}{27} \times B_C \\ \Rightarrow B C & =27 B \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Torque, } \tau=N L A B \sin \theta \\ & =200 \times 3 \times 10^{-3} \times\left(10 \times 10^{-2}\right)^2 \times 2 \times \sin 90^{\circ} \\ & =12 \times 10^{-3} \mathrm{~N}-\mathrm{m} \end{aligned} $

The length of the wire $L$ is constant for both coils

For coil $A, L=N_A \times 2 \pi R_A$

For coil $B, L=N_B \times 2 \pi R_B$

$ \begin{aligned} & N_A \times 2 \pi R_A=N_B \times 2 \pi R_B \\ & N_A R_A=N_B R_B \end{aligned} $

So, $\quad \frac{R_A}{R_B}=\frac{N_B}{N_A}$

Magnetic field at the centre of $\operatorname{coil} A$

$ B_A=\frac{\mu_0 N_A I}{2 R_A} $

Magnetic field at the centre of coil $B$

$ B_B=\frac{\mu_0 N_B I}{2 R_B} $

The ratio is $\frac{B_A}{B_B}=\frac{\frac{\mu_0 N_A I}{2 R_A}}{\frac{\mu_0 N_B I}{2 R_B}}$

$ \begin{aligned} & \frac{B_A}{B_B}=\frac{N_A}{N_B} \times \frac{R_B}{R_A}=\left(\frac{N_A}{N_B}\right) \cdot\left(\frac{N_A}{N_B}\right)=\left(\frac{N_A}{N_B}\right)^2 \\ & \frac{B_A}{B_B}=\left(\frac{2}{3}\right)^2=\frac{4}{9} \end{aligned} $

The ratio of the magnetic fields at the centres of coils $A$ and $B$ is $4: 9$.

$ \begin{array}{ll} & V_{A D}=I_1 \cdot R_{A B C D}=I_2 R_{A D} \\ \Rightarrow \quad & I_1 \times 3 R=I_2 \times R \Rightarrow I_2=3 I_1 \end{array} $

Also $I_1+I_2=4 \mathrm{~A}$      (given)

$\Rightarrow \quad 4 I_1=4 \mathrm{~A}$ or $I_1=1 \mathrm{~A}$

So, we have following situation,

$\therefore$ Magnetic field $B$ at centre

$ =B_{A B}+B_{B C}+B_{C D}-B_{A D} $

Here, $B_{A B}=B_{B C}=B_{C D}$

$ \begin{aligned} & =\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \times 2 \sin \theta     \,\,\,\,\,\left[\theta=45^{\circ}\right]\\ & =\frac{\mu_0}{4 \pi} \cdot \frac{1}{2.5 \times 10^{-2}} \times \frac{2}{\sqrt{2}} \\ & =\frac{10^{-7} \times \sqrt{2}}{25 \times 10^{-2}}=4 \sqrt{2} \times 10^{-6} \mathrm{~T} \end{aligned} $

$ \begin{aligned} \text { And } B_{A D} & =3 \times 4 \sqrt{2} \times 10^{-6} \\ & =12 \sqrt{2} \times 10^{-6} \mathrm{~T} \\ \therefore \quad B_{\text {centre }} & =3 \times\left(4 \sqrt{2} \times 10^{-6}\right)-\left(12 \sqrt{2} \times 10^{-6}\right) \\ & =0 \end{aligned} $

When a charged particle moves in a magnetic field, the time it takes to make one full round (the time period) 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. This means the time period depends on the ratio of mass to charge ($ \frac{m}{q} $).

To compare the time periods of a proton ($ T_p $) and an alpha particle ($ T_\alpha $), use this ratio:

$ \frac{T_p}{T_\alpha} = \frac{m_p}{m_\alpha} \times \frac{q_\alpha}{q_p} $

An alpha particle has a mass four times that of a proton ($ m_\alpha = 4m_p $) and a charge twice that of a proton ($ q_\alpha = 2q_p $).

Plug these values in:

$ \frac{T_p}{T_\alpha} = \frac{m_p}{4m_p} \times \frac{2q_p}{q_p} = \frac{1}{4} \times 2 = \frac{1}{2} $

This shows that the time taken by a proton is half that of the alpha particle. So, the ratio of their times is

$ T_p : T_\alpha = 1 : 2 $

$ \begin{aligned} &\text { Number of turns per unit length }\\ &\begin{aligned} n & =\frac{N}{L}=\frac{200}{0.5}=400 \text { turns } / \mathrm{m} \\ \therefore \quad B & =\mu_0 n I \end{aligned} \end{aligned} $

$ \begin{aligned} & =4 \pi \times 10^{-7} \times 400 \times 2.5 \\ & =4 \pi \times 10^{-4} \mathrm{~T} \end{aligned} $

$ \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

Given,

The $ \alpha $-particle is travelling along the $ x $-direction, $ \mathbf{v}=v \hat{\mathbf{i}} $

The magnetic field is applied in the $ y $-direction, $ \mathbf{B}=B \hat{\mathbf{j}} $

Thus, Lorentz force,

$ \begin{array}{l} \mathbf{F}=q(\mathbf{v} \times \mathbf{B}) \\\\ \mathbf{F}=2 \cdot e(v \hat{\mathbf{i}}) \times(\hat{B \mathbf{j}}) \end{array} $

$ \begin{array}{l} \mathbf{F}=2 e v B \hat{\mathbf{k}} \quad[\therefore \hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}}] \end{array} $

The force is acting in the $ +z $-direction.

This force is always perpendicular to both the velocity of the $ \alpha $-particle and the magnetic field. It will cause the $ \alpha $-particle to move in a circular path. Therefore, the motion of the $ \alpha $-particle will be a circular path in the $ X Z $-plane, perpendicular to the direction of the magnetic field.

Given, $ B=3 \mathrm{~T} $

$ \begin{array}{l} i=2 \sqrt{2} \mathrm{~A} \\\\ \theta=45^{\circ} \end{array} $

The force $ F $ on a current carrying conductor is given by,

$ F=I l B \sin \theta $

Force per unit length,

$ \begin{array}{l} \frac{F}{l}=I B \sin \theta \\\\ \frac{F}{l}=2 \sqrt{2} \times 3 \times \sin 45 \\\\ \frac{F}{l}=2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}} \\\\ \frac{F}{l}=6 \mathrm{Nm}^{-1} \end{array} $

Given,

Velocity of electron is downward.

Magnetic field along south.

As from Lorentz equation,

$ \mathbf{F}_{\text {Magnetic }}=q(\mathbf{v} \cdot \mathbf{B}) $

Therefore the direction of $ \mathbf{F} $ will be along $ (\mathbf{v} \times \mathbf{B}) $.

Hence, by using right hand thumb rule for negative charge, the direction will be east.

Current in wire $ A $

$ \Rightarrow \quad I_{A}=2 \mathrm{~A} $

Current in wire $ B $

$ I_{B}=6 \mathrm{~A} $

We have to find net magnetic field at point $ P $.

Magnetic field due to wire $ A $ will be inward.

Magnetic field due to wire $ B $ will be outward.

$ B_{\mathrm{Net}}=B_{B}-B_{A} $

Using Ampare's circuital law

$ \begin{array}{l} B_{A}=\frac{\mu_{0} i_{A}}{2 \pi R_{A}} \\\\ B_{A}=\frac{\mu_{0}(2)}{2 \pi(2)}=\frac{\mu_{0}}{2 \pi} \\\\ B_{B}=\frac{\mu_{0}(6)}{2 \pi \dot{\times} 3}=\frac{\mu_{0}}{\pi} \\\\ B_{\text {Net }}=\frac{\mu_{0}}{\pi}-\frac{\mu_{0}}{2 \pi}=\frac{\mu_{0}}{2 \pi} \\\\ \mu_{0}=4 \pi \times 10^{-7} \\\\ B_{\text {Net }}=\frac{4 \pi \times 10^{-7}}{2 \pi} \end{array} $

$ B_{\text {Net }} $ at point $ P=2 \times 10^{-7} \mathrm{~T} $

Given,

Diameter $ =4 \mathrm{~mm} $

Current in wire $ =i $

Magnetic field inside the pipe

Using Ampere's circuital law for $ P_{1} $ loop

$ \begin{aligned} \oint \mathbf{B}_{1} d \mathbf{L} & =\mu_{0} I_{\text {enclosed }} \\\\ B_{1} & =\frac{\mu_{0} i_{1}}{2 \pi r^{2}} \quad\left[r_{1}=1 \mathrm{~mm}\right] \\\\ B_{1} & =\frac{\mu_{0} i}{8 \pi} \quad[r=2 \mathrm{~mm} \text { of wire }] \end{aligned} $

For point $ P_{2} $

$ \begin{aligned} \oint \mathbf{B}_{2} d \mathbf{L} & =\mu_{0} I_{\text {enclosed }} \\\\ B_{2} & =\frac{\mu_{0} i}{2 \pi \times\left(r_{2}\right)} \quad\left[r_{2}=4 \mathrm{~mm}\right] \\\\ B_{2} & =\frac{\mu_{0}}{8 \pi} i \end{aligned} $

From Eqs. (i) and (ii)

$ \frac{B_{1}}{B_{2}}=\frac{1}{1}=1: 1 $

Given,

Straight wire length, $ l=20 \mathrm{~cm} $

Curent $ I=\frac{3}{\pi^{2}} \mathrm{~A} $

Length of wire $ = $ Circumference of circle

$ 20 \mathrm{~cm}=2 \pi r $

$ r=\frac{20}{2 \pi} \mathrm{~cm} \text { or } \frac{20 \times 10^{-2}}{2 \pi} \mathrm{~m} $

Magnetic field at the centre of circle

$ \begin{aligned} B & =\frac{\mu_{0} I}{2 R} \\\\ & =\frac{4 \pi \times 10^{-7} \times 3}{2 \times \frac{20 \times 10^{-2}}{2 \pi} \times \pi^{2}} \\\\ & =\frac{4 \pi \times 3 \times 2 \pi \times 10^{-7}}{2 \times 20 \times \pi^{2} \times 10^{-2}} \\\\ & =6 \times 10^{-6} \mathrm{~T} \end{aligned} $

Initial current in loop $ I_{1}=2.5 \mathrm{~A} $

Let $ m $ is magnetic moment of loop, $ B $ is external magnetc field.

Then time period of oscillation,

$ T=\frac{1}{2 \pi} \sqrt{\frac{m B}{I}} $

or

$ T=\frac{1}{2 \pi} \sqrt{\frac{m B}{2.5}} $

Now, when current is 10 A

$ T^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{m B}{10}} $

or

$ \frac{1}{2 \pi} \sqrt{\frac{m B}{4 \times 2.5}} $

or

$ \frac{1}{2} \times \frac{1}{2 \pi} \sqrt{\frac{m B}{2.5}} $

Hence, $ T^{\prime}=T / 2 $

The magnetic field $B$ is the circular coil is given as

$ B=\frac{\mu_0 I r^2}{2\left(r^2+x^2\right)^{3 / 2}} $

Given, $x_1=4 \mathrm{~cm}$ and $x_2=3 \sqrt{3} \mathrm{~cm}$

According to question,

$ \begin{aligned} & B_1=\frac{\mu_0 I r^2}{2\left(r^2+4^2\right)^{3 / 2}} \quad \ldots \text { (i) } \\\\ & B_2=\frac{\mu_0 I r^2}{2\left[r^2+(3 \sqrt{3})^2\right]^{3 / 2}} \quad \ldots \text { (ii) } \end{aligned} $

On dividing Eqs. (i) by (ii), we get

$ \begin{aligned} & \therefore \quad \frac{216}{125}=\frac{\mu_0 I r^2}{2\left(r^2+16\right)^{3 / 2}} \times \frac{2\left(r^2+27\right)^{3 / 2}}{\mu_0 I r^2} \\\\ & \Rightarrow \quad\left(\frac{6}{5}\right)^3=\frac{\left(r^2+27\right)^{3 / 2}}{\left(r^2+16\right)^{3 / 2}} \\\\ & \Rightarrow \quad\left(\frac{6}{5}\right)=\frac{\left(r^2+27\right)^{\frac{1}{2}}}{\left(r^2+16\right)^{\frac{1}{2}}} \\\\ & \Rightarrow \quad \frac{36}{25}=\frac{r^2+27}{r^2+16} \\\\ & \Rightarrow \quad 36\left(r^2+16\right)=25\left(r^2+27\right) \\\\ & \Rightarrow \quad 36 r^2+576=25 r^2+675 \\\\ & \Rightarrow \quad 11 r^2=99 \\\\ & \Rightarrow \quad r^2=9 \\\\ & \Rightarrow \quad r=3 \mathrm{~cm} \end{aligned} $

The pitch of a charged particle having charge $q$ and mass $m$ is given by

$ \begin{aligned} r & =\frac{2 \pi m \times v \sin \theta}{q B} \\\\ \therefore \quad \frac{r_1}{r_2} & =\frac{2 \pi m_1 \times v \sin \theta}{q_1 B} \times \frac{q_2 B}{2 \pi m_2 v \sin \theta} \\\\ & =\frac{q_2 m_1}{q_1 m_2} \\\\ & =\frac{3 q \times m}{2 q \times 2 m} \Rightarrow \frac{3}{4} \end{aligned} $

Charge, $q=2 \mathrm{C}$

Velocity,

$ v=(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s} $

Magnetic field,

$ \mathbf{B}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \mathrm{T} $

Electric field,

$ \mathbf{E}=(-2 \hat{\mathbf{k}}) N C^{-1} $

Lorentz force is given by

$ \begin{aligned} & \mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B}) \\\\ & \mathbf{v} \times \mathbf{B}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ 3 & 4 & 0 \\\\ 1 & 2 & 3 \end{array}\right| \\\\ & =(12-0) \hat{\mathbf{i}}-(9-0) \hat{\mathbf{j}}+(6-4) \hat{\mathbf{k}} \\\\ & \mathbf{v} \times \mathbf{B}=12 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\\\ & \text { Thus, } \mathbf{F}=2(-2 \hat{\mathbf{k}})+2(12 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\\\ & \mathbf{F} \\\\ & =24 \hat{\mathbf{i}}-18 \hat{\mathbf{j}}+(4-4) \hat{\mathbf{k}} \\\\ & \mathbf{F} \end{aligned} $

Magnitude of Lorentz force,

$ \begin{aligned} |\mathbf{F}| & =\sqrt{(24)^2+(-18)^2} \\\\ & =\sqrt{576+324} \\\\ & =\sqrt{900} \\\\ |\mathbf{F}| & =30 \mathrm{~N} \end{aligned} $

$ \begin{aligned} & N=400 \text { turns } \\\\ & A=10^{-2} \mathrm{~m}^2, I=0.5 \mathrm{~A} \\\\ & B=1 \mathrm{~T}, \theta=\left(90^{\circ}-60^{\circ}\right)=30^{\circ} \end{aligned} $

As we know, $\tau=N I B A \sin \theta$

$ \begin{aligned} \tau & =400 \times 0.5 \times 1 \times 10^{-2} \times \sin 30^{\circ} \\\\ \tau & =400 \times 0.5 \times 10^{-2} \times \frac{1}{2} \\\\ \tau & =200 \times 0.5 \times 10^{-2} \\\\ & =20 \times 5 \times 10^{-2}=100 \times 10^{-2} \\\\ \tau & \cong 1 \mathrm{~N}-\mathrm{m} \end{aligned} $

The initial moment of force acting on the coil in Nm is 1.

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