Moving Charges and Magnetism

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) $

Given,

Square loop with side length $=a$

Current in loop $=I$

Magnetic field = $B$

The Net force is given by Lorentz equation, Initially when direction of current clockwise.

$ \begin{aligned} F_{\mathrm{net}} & =F_g+B I L \sin \theta \\ & =m g+I a B\left[L=a, \sin 90^{\circ}=1\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

When direction of current is reversed,

$ \begin{aligned} F_{\mathrm{net}} & =F_g+B I L \sin \theta \\ & =m g-I a B\left[L=a, \sin 180^{\circ}=-1\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii)

$ \begin{aligned} & F_{\mathrm{net}}=F_1-F_2 \\ & \Delta F_{\mathrm{net}}=2 \mathrm{laB} \end{aligned} $

Given,

Current is circulating in loop $=I$

Magnetic field $=B$

Potential energy of a current carrying loop when it placed in a magnetic field is given by

$U=-\mu B \cos \theta[\mu$ is magnetic moment $\mu= \pm A]$

$ =-I A B \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For loop (1), $\theta=180^{\circ}$

$ U_1=-I A B(-1)=+I A B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

For loop (2), $\cos \theta=90$

$ U_2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) $

For loop (3), $\theta<90$

$ U_3=-I A B \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

For loop (4), $\theta=(180>\theta>90)$

$ \begin{aligned} & =+\mathrm{ve} \text { in second quadrant } \\ U_4 & =+I A B \cos \theta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{aligned} $

Hence correct order is

$ U_1>U_4>U_2>U_3 $

Option (c) is the correct option

Given,

Current in wire $=i$

Wire length $=L$

When the wire of length $L$ is bent in circular shape of radius $r$.

$ r=\frac{L}{2 \pi} $

Area of circular loop $=\pi r^2$

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

Magnetic moment is given by

$ \begin{aligned} & M=I A=i \times \pi\left(\frac{L}{2 \pi}\right)^2 \\ & M=\frac{L^2 i}{4 \pi} \end{aligned} $

Hence, option (a) is correct option.

Given, Magnetic field $B=0$ Let $I$ be the non zero current. $d s$ is the length element.

$r$ is the separation distance between wire element. Using Biot-Savart's Law

$ \begin{aligned} |d B| & =\frac{\mu_0}{4 \pi} \frac{I d l \sin \theta}{r^2} \\ \text { or } \quad \mathbf{B} & =\frac{\mu_0}{4 \pi} \frac{I L \sin \theta \hat{\mathbf{r}}}{r^2} \end{aligned} $

[ $L$ is total length of current carrying element]

$ B=0 $            [given]

For $B$ to be zero, either I should be zero or $\sin \theta$ be zero.

But in question, it is mention current is non zero.

$ \therefore \quad \sin \theta=0 \quad \theta=0, \pi $

So, the angle between current and position vector is $0^{\circ}$ or $180^{\circ}$.

Given

Force per unit length on $C$ is $F$.

Calculating net force on $C$.

$ \frac{F_{\text {net }}}{L}=\frac{F_{C A}}{L}+\frac{F_{C B}}{L} $

$ F \text { is given by } \frac{F}{L}=\frac{\mu_0}{4 \pi} \frac{2 i \times i_2}{d} $

Where $d$ is separation distance $F$ due to $A$

$ \frac{F}{L}=-\frac{\mu_0}{4 \pi} \frac{2 \times 3 i \times 2 i}{d}=+\frac{\mu_0}{4 \pi} \times \frac{12 i^2}{2 d} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$F$ on $C$ due to $B$

$ \begin{gathered} \frac{F}{L}=\frac{\mu_0}{4 \pi} \frac{2 \times 2 i \times i}{d} \\ =-\frac{\mu_0}{4 \pi} \frac{4 i^2}{d} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \frac{F_{\mathrm{net}}}{L}=+\frac{\mu_0 i^2}{2 \pi d}=F \text { (given) } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

Now $B$ is removed, Net Force is (on $A$ )

$ \frac{F_{(A C)}}{L}=-\frac{\mu_0}{4 \pi}\left[\frac{6 i \times 2 i}{2 d}\right] \Rightarrow-3\left[\frac{\mu_0 i^2}{2 \pi d}\right]=-3 F $

Using Ampere's circuital law over a circular loop of any radius less than the radius of the pipe, we can see that net current inside the pipe is zero. Hence, megnetic field at every point inside the pipe is zero.

Given Length of solenoid,

$ \begin{aligned} & l=80 \mathrm{~cm}=0.8 \mathrm{~m} \\ & I=8 \mathrm{~A} \end{aligned} $

Total number of turns,

$N=$ No. of layers × No. of turns in each layer

$ =5 \times 400=2000 $

$ \begin{aligned} &\text { ∴ No. of turns per unit length, }\\ &\begin{aligned} n & =\frac{N}{l} \\ & =\frac{2000}{0.8}=2500 \mathrm{turns} / \mathrm{m} \\ B & =\mu_0 n I \\ & =4 \pi \times 10^{-7} \times 2500 \times 8 \\ & =2.5 \times 10^{-2} \mathrm{~T} \end{aligned} \end{aligned} $

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

$ \begin{array}{rlrl} & v =2 \mathrm{~m} / \mathrm{s} \text { along } X \text {-axis } \\ \therefore \quad \mathbf{v} & =2 \hat{\mathbf{i}} \mathrm{~m} / \mathrm{s} \end{array} $

Force on charged particle,

$ \begin{aligned} \mathbf{f} & =q(\mathbf{v} \times \mathbf{B}) \\ & =q[2 \hat{\mathbf{i}} \times(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})] \\ & =q[0+4 \hat{\mathbf{k}}-6 \hat{\mathbf{j}}] \\ & =q(4 \hat{\mathbf{k}}-6 \hat{\mathbf{j}}) \\ & =q(-6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \end{aligned} $

From the above expression of force, it is clear that charge will experience a force in $Y Z$ Plane.

To determine the magnetic field at a distance of one meter from a wire carrying a current of 1 ampere, we apply the Biot-Savart law.

Given:

Distance, $ r = 1 \, \text{m} $

Current, $ I = 1 \, \text{A} $

The magnetic field at a perpendicular distance from a straight current-carrying wire can be calculated using the formula derived from Biot-Savart's Law:

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

where $ \mu_0 = 4 \pi \times 10^{-7} \, \mathrm{H/m} $ is the permeability of free space.

Calculation:

$ B = \frac{4 \pi \times 10^{-7}}{2 \pi \times 1} \times 1 = 2 \times 10^{-7} \, \mathrm{T} $

Therefore, the magnetic field at a distance of one meter from the wire is $ 2 \times 10^{-7} \, \mathrm{T} $.

Given:

Number of turns, $ N = 1000 $

Area of the coil, $ A = 2 \, \text{cm}^2 $ or $ 2 \times 10^{-4} \, \text{m}^2 $

Current through the coil, $ I = 1 \, \text{A} $

The magnetic moment ($ M $) of a coil is calculated using the formula:

$ M = N \times I \times A $

Substitute the given values:

$ M = 1000 \times 1 \times 2 \times 10^{-4} $

$ M = 0.2 \, \text{A} \cdot \text{m}^2 $

Therefore, the magnetic moment of the coil is $ 0.2 \, \text{A} \cdot \text{m}^2 $.

Given,

Wire $P Q$ carries current $=10 \mathrm{~A}$

Magnetic field (uniform) $=5 \mathrm{~T}$

Which is normal to plane of paper.

Let $P Q$ be a straight wire, then

$ \left[\triangle P B O, P O^2=P B^2+B V^2\right] $

$ \text { Length of } \begin{aligned} P Q & =P O+O Q \\ & =5+5=10 \mathrm{~cm} \end{aligned} $

Force experienced by wire in uniform magnetic field is given by

$ F=B I L \sin \theta $

Here, $\theta=90^{\circ}$

[Plane and outward (normal) to plane]

$ \begin{aligned} F & =5 \times 10 \times \sin 90^{\circ} \times(0.1 \mathrm{~m}) \\ \Rightarrow \quad F & =5 \mathrm{~N} \end{aligned} $

Given:

Current in the long straight wire, $ I = 18 \, \mathrm{A} $

Distance from the wire where the magnetic field is measured, $ r = 12 \, \mathrm{cm} = 0.12 \, \mathrm{m} $

To find the magnetic field using Ampere's law for a long straight wire:

$ \oint B \cdot dl = \mu_0 I $

This simplifies to:

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

Substituting the known values into the equation:

$ B = \frac{(4 \pi \times 10^{-7}) \times 18}{2 \pi \times 0.12} $

Simplifying, we get:

$ B = 3 \times 10^{-5} \, \mathrm{T} $

The magnitude of the magnetic field at a point 12 cm from the wire is $ 3 \times 10^{-5} \, \mathrm{T} $.

Magnetic field at the origin $O$ due to both Wire-1 and Wire -2 is given as

$ \begin{aligned} B=B_1+ & B_2 \\ & =\frac{\mu_0}{2 \pi} \times \frac{i}{r_1}+\frac{\mu_0}{2 \pi} \times \frac{i}{r_2} \\ & =2 \times 10^{-7} i\left[\frac{1}{r_1}+\frac{1}{r_2}\right] \\ & =2 \times 10^{-7} i\left[\frac{1}{1 \times 10^{-2}}+\frac{1}{2 \times 10^{-2}}\right] \\ & =2 \times 10^{-7} i[150]=3 \times 10^{-5} i \end{aligned} $

∴ Magnetic field at $O$ due to Wire- 1 alone

$ \begin{aligned} B_0 & =\frac{\mu_0}{2 \pi} \times \frac{i}{r_1} \\ & =2 \times 10^{-7} \frac{i}{1 \times 10^{-2}}=2 \times 10^{-5} i \\ \therefore \quad \frac{B}{B_0} & =\frac{3 \times 10^{-5} i}{2 \times 10^{-5} i}=\frac{3}{2} \end{aligned} $

Given, inner radius and outer radius of toroid,

$ r_1=0.24, \mathrm{~cm}, r_2=0.26 \mathrm{~m} $

Current, $I=10 \mathrm{~A}$

Number of turns, $N=2500$

Number of turns per unit length of toroid,

$ \begin{aligned} n & =\frac{N}{2 \pi\left(\frac{r_1+r_2}{2}\right)} \\ & =\frac{2500}{2 \pi\left(\frac{0.24+0.26}{2}\right)} \\ & =\frac{2500}{2 \pi \times 0.25}=\frac{1 \times 10^4}{2 \pi} \end{aligned} $

∴ Magnetic field inside the core of toroid,

$ \begin{aligned} B & =\mu_0 n I \\ & =4 \pi \times 10^{-7} \times \frac{1 \times 10^4}{2 \pi} \times 10 \\ & =2 \times 10^{-2} \mathrm{~T} \end{aligned} $

Current carrying loop $A B C D$ is shown below

Current, $I=\frac{1.2}{\pi} \mathrm{~A}$

Magnetic field at centre $O$ due to current carrying arc $A D$,

$ \begin{aligned} B_1 & =\frac{\mu_0 I}{2 r_1} \cdot \frac{\theta}{360^{\circ}} \\ & =\frac{4 \pi \times 10^{-7} \times \frac{1.2}{\pi}}{2 \times 1 \times 10^{-2}} \times \frac{30^{\circ}}{360^{\circ}} \end{aligned} $

$ \begin{aligned} & =24 \times 10^{-5} \times \frac{1}{12} \\ & =2 \times 10^{-6} \mathrm{~T} \text { (upward direction) } \end{aligned} $

Similarly, magnetic field at centre $O$ due to current carrying arc $B C$,

$ \begin{aligned} B_2 & =\frac{\mu_0 I}{2 r_2} \cdot \frac{\theta}{360^{\circ}} \\ & =\frac{4 \pi \times 10^{-7} \times \frac{1.2}{\pi}}{2 \times 2 \times 10^{-2}} \times \frac{30}{360^{\circ}} \\ & =1 \times 10^{-6} \mathrm{~T} \quad \text { (downward direction) } \end{aligned} $

∴ Net magnetic field at centre $O$,

$ \begin{aligned} B_{\text {net }} & =B_1-B_2=2 \times 10^{-6}-1 \times 10^{-6} \\ & =1 \times 10^{-6} \mathrm{~T}=1 \mu \mathrm{~T}(\text { upward }) \end{aligned} $

The given situation is shown below

Magnitude of force on a length $l$ of wire $a$ due to current carrying parallel wire $b$,

$ F_{a b}=\frac{\mu_0}{2 \pi} \cdot \frac{i_a i_b}{d_1} \times l \text { (Attractive) } $

Similarly, magnitude of force on a length $l$ of wire $a$ due to current carrying wire $c$,

$ F_{a c}=\frac{\mu_0}{2 \pi} \frac{i_a i_c}{d_2} \times l \text { (Repulsive) } $

∴ Net magnetic force on a length of the wire $a$,

$ \begin{aligned} F_{\mathrm{net}} & =F_{a c}-F_{a b} \\ & =\frac{\mu_0}{2 \pi} \frac{i_a i_c}{d_2} \times l-\frac{\mu_0}{2 \pi} \frac{i_a i_b}{d_1} \times l \\ & =\frac{\mu_0}{2 \pi} \cdot \frac{i_a \times 4 i_a \times l}{2 d_1}-\frac{\mu_0}{2 \pi} \cdot \frac{i_a \times i_a \times l}{d_1} \\ & =\frac{\mu_0 i_a^2 l}{2 \pi d_1} \end{aligned} $

(since, $d_2=2 d_1, i_b=i_a$ and $i_c=4 i_a$ )

When an electron is projected in the direction of electric and magnetic field, magnetic field does not produce any effect on the velocity of electron because no force acts on the electron when it is moving in the direction of magnetic field. But an electric force acts on the electron in opposite direction of electric field, hence electron velocity will decrease in magnitude. Therefore, statement I is true.

Two infinite long parallel current carrying wires having equal current in the same direction produce equal value of magnetic field but in opposite direction at the mid-way point between the wires, hence net magnetic field at this point is zero.

Length of each parallel conductor,

$ l=50 \mathrm{~m} $

Separation between the both conductors,

$ r=0.2 \mathrm{~m} $

Force between both conductor,

$ F=1 \mathrm{~N} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & & F & =\frac{\mu_0}{2 \pi} \frac{I_1 I_2}{r} l \\ &\Rightarrow & 1 & =\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{I_1 I_2 \times 50}{0.2} \\ &\Rightarrow & 1 & =5 I_1 I_2 \times 10^{-5} \\ & \Rightarrow & I_1 I_2 & =\frac{1}{5 \times 10^{-5}} \\ & \Rightarrow & I_1 I_2 & =2 \times 10^4 \\ & \Rightarrow & 2 I_2 \cdot I_2 & =2 \times 10^4 \quad \quad\left[\text { Given, } I_1=2 I_2\right] \\ & \Rightarrow & I_2 & =10^2 \mathrm{~A}=100 \mathrm{~A} \end{aligned} \end{aligned} $

At a distance of 5 cm from axis point at which field is observed is $P$.

At $P$, two perpendicular fields exists. $B_1$ is magnetic field of solenoid and $B_2$ is magnetic field of wire.

Now, $B_1=\mu_0 n I_1$

$ =\mu_0(1000) \times 7 \times 10^{-3}=7 \mu_0 \mathrm{~T} $

And $B_2=\frac{\mu_0 I_2}{2 \pi d}=\frac{m_0 I_2}{2 \pi \times 5 \times 10^{-2}}=\frac{10 \mu_0 I_2}{\pi} \mathrm{~T}$

As, $B_1=B \cos \theta$

and $B_2=B \sin \theta$

$ \begin{aligned} & \Rightarrow \quad \tan \theta=\frac{B_2}{B_1} \\ & \Rightarrow \quad B_2=B_1 \tan 60^{\circ}(\text { Given, } \theta=60 \% \end{aligned} $

Substituting values of $B_2$ and $B_1$, we get

$ \begin{aligned} & & \frac{10 \mu_0 I_2}{\pi} & =7 \mu_0 \times \sqrt{3} \\ \Rightarrow & & I_2 & =\frac{7 \times \pi \times \sqrt{3}}{10} \mathrm{~A} \\ \Rightarrow & & I_2 & =3.74 \mathrm{~A} \end{aligned} $

Since, wire hangs in air, hence

$ \begin{gathered} m g=I B l \Rightarrow \frac{m}{l}=\frac{I B}{g} \\ \Rightarrow \quad \frac{m}{l}=\frac{I}{g} \cdot \frac{\mu_0 I}{2 \pi r} \quad\left(\because B=\frac{\mu_0 I}{2 \pi r}\right) \end{gathered} $

Here, $I=160 \mathrm{~A}, \frac{m}{l}=10 \mathrm{~g} / \mathrm{m}=10^{-2} \mathrm{~g} / \mathrm{m}$

$ \begin{aligned} r & =4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~cm} \\ g & =10 \mathrm{~m} / \mathrm{s}^2, \mu_0=4 \pi \times 10^{-7} \\ \therefore 10^{-2} & =\frac{160}{10} \times \frac{4 \pi \times 10^{-7} \times I}{2 \pi \times 4 \times 10^{-2}} \\ \Rightarrow 10^{-2} & =8 \times 10^{-5} \times I \Rightarrow I=\frac{10^3}{8}=125 \mathrm{~A} \end{aligned} $

Number of turns per unit length of solenoid, $n=70$ turns/ $\mathrm{cm}=7000$ turns/m

∴ Magnetic field inside the solenoid due to current $I$,

$ B=\mu_0 n I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Since, electron is moving inside the solenoid in circular path of radius $r$, then $r=m v / B e$

$ \begin{array}{ll} \Rightarrow & r=\frac{m v}{\mu_0 n l e}\,\,\,\,\,[From Eq. (i)] \\ \Rightarrow & I=\frac{m v}{\mu_0 n r e} \end{array} $

Here, $m=9.1 \times 10^{-31} \mathrm{~kg}, v=4.4 \times 10^6 \mathrm{~m} / \mathrm{s}$

$ \begin{aligned} e & =1.6 \times 10^{-19} \mathrm{C} \\ \Rightarrow \quad \mu_0 & =4 \pi \times 10^{-7} \mathrm{SI} \text { unit } \\ r & =2.5 \mathrm{~cm} \\ & =2.5 \times 10^{-2} \mathrm{~m} \end{aligned} $

Putting these values in the above equation, we get

$ \begin{array}{r} I=\frac{9.1 \times 10^{-31} \times 4.4 \times 10^6}{4 \pi \times 10^{-7} \times 7000 \times 2.5 \times 10^{-2}} \\ \times 1.6 \times 10^{-19} \\ =0.1125 \mathrm{~A}=112.5 \mathrm{~mA} \end{array} $

The given situation is shown as

Magnitude of magnetic field due to straight current carrying wire $A B$ is given as

$ \begin{aligned} B_1= & \frac{\mu_0}{4 \pi} \cdot \frac{I}{O M}(\sin \phi+\sin \phi) \\ = & \frac{\mu_0}{4 \pi} \times \frac{I}{R \sin \phi}(2 \sin \phi) \quad(\because O M=R \sin \phi) \\ = & \frac{\mu_0}{4 \pi} \times \frac{2 I}{R}=10^{-7} \times \frac{2 \times 5}{0.1} \\ & \quad(\because R=100 \mathrm{~mm}=0.1 \mathrm{~m} \text { and } I=5 \mathrm{~A}) \\ = & 10^{-5} \mathrm{~T} \end{aligned} $

Magnitude of magnetic field due to current carrying circular arc,

$ \begin{aligned} B_1 & =\frac{\mu_0 I}{2 R} \times \frac{(2 \pi-2 \phi)}{2 \pi} \\ & =\frac{4 \pi \times 10^{-7} \times 5}{2 \times 0.1} \times \frac{\left(2 \pi-\frac{\pi}{2}\right)}{2 \pi} \\ & \left(\because 2 \phi=90^{\circ}=\frac{\pi}{2}\right) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{4 \times 314 \times 10^{-7} \times 5}{0.2} \times \frac{3}{4} \\ B_2 & =2.36 \times 10^{-5} \mathrm{~T} \end{aligned}\\ &\text { ∴ Total magnetic field at } O \text {, }\\ &\begin{aligned} B & =B_1+B_2=10^{-5}+2.36 \times 10^{-5} \\ & =10^{-5}(1+2.36)=3.36 \times 10^{-5} \mathrm{~T} \\ & =33.6 \times 10^{-6} \mathrm{~T}=33.6 \mu \mathrm{~T} \end{aligned} \end{aligned} $

Given, inner radius of toroid,

$ r_1=24 \mathrm{~cm}=0.24 \mathrm{~m}, $

Outer radius, $r_2=26 \mathrm{~cm}=0.26 \mathrm{~m}$

Current, $I=12 \mathrm{~A}, N=2000$ turns

Magnetic field inside the core of toroid,

$ B=\left(\frac{\mu_0 N I}{l}\right) $

where, $l=2 \pi\left(\frac{r_1+r_2}{2}\right)=2 \pi\left(\frac{0.24+0.26}{2}\right)$

$ \begin{aligned} & =2 \pi \times 0.25=0.5 \pi \\ \therefore \quad B & =\frac{\mu_0 N I}{0.5 \pi} \\ & =\frac{4 \pi \times 10^{-7} \times 2000 \times 12}{0.5 \pi} \\ & =1.92 \times 10^{-2} \mathrm{~T} \end{aligned} $

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.

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

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

As, we know that, force on a moving charge particle.

$\therefore F=Bqv\sin\theta$

Hence, F will be zero, if any of the parameter of above formula equals zero.

$\therefore$ Force on a charge at rest will be zero.

Given, radius of hollow cylinder = R

Current = i

Distance from axis = r

Let magnetic field be B.

$\because$ No current is passing through region r < R

and magnetic field $(B)=\frac{1}{2\pi}\frac{\propto_0I}{r}$

$\begin{aligned} \text{So,} \\& B(r < R)=0 \\ & B(r=R)=\frac{1}{2 \pi} \frac{\propto_0 I}{R} \\ \text {and}\\& B(r>R)=\frac{1}{2 \pi} \frac{\propto_0 I}{r} \end{aligned}$

So, these variations are correctly shown as below.

Given, mass of the wire,

$ m=0.2 \mathrm{~kg} $

Length of the wire, $l=1.5 \mathrm{~m}$

Current, $I=2 \mathrm{~A}$

The force acting on the current carrying wire in uniform magnetic field,

$ \begin{aligned} & F=I B l \sin \theta \\ & F=I B l \sin 90^{\circ} \quad\left(\because \theta=90^{\circ}\right) \end{aligned} $

$ \begin{equation*} F=I B l \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Weight of the wire, $w=m g=0.2 \times 10 \mathrm{~N}$ In the position of suspension,

$ B I l=m g \Rightarrow B=\frac{m g}{I l}=\frac{0.2 \times 10}{2 \times 1.5}=0.67 \mathrm{~T} $

The magnetic field at some internal point on the axis of a solenoid is given by

$ B=\frac{\mu_0 n I}{2}\left(\cos \theta_1-\cos \theta_2\right) $

Magnetic field at the centre of a long solenoid is given by setting $\theta_1=0^{\circ}$ and $\theta_2=\pi$.

$ \begin{align*} \Rightarrow \quad B_1 & =\frac{\mu_0 n I}{2}\left(\cos 0^{\circ}-\cos \pi\right) \\ & =\frac{\mu_0 n I}{2}[1-(-1)] \\ & =\mu_0 n I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{align*} $

Similarly, the magnetic field at the end of a long solenoid is given by setting $\theta_1=\frac{\pi}{2}$ and $\theta_2=\pi$.

$ \begin{align*} \Rightarrow \quad B_2 & =\frac{\mu_0 n I}{2}\left(\cos \frac{\pi}{2}-\cos \pi\right) \\ & =\frac{\mu_0 n I}{2}[0-(-1)] \\ & =\frac{\mu_0 n I}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

From Eq. (i) and Eq. (ii), we get

$ \frac{B_1}{B_2}=\frac{\mu_0 n I}{\mu_0 n I} \times 2=2 $