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} $

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.

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} $

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 $

The magnetic field at point due to a current carrying straight conductor of infinite length is

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

The magnetic flux through circular loop around conductor in $x y$-plane is

$ \phi_B=B A=\frac{\mu_0 I}{2 \pi R} \times \pi R^2=\frac{\mu_0 I R}{2} $

The given situation is shown in the following figure

Magnetic field at point $P$ due to the wire having current $I$ is given as

$ B_I=\frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y} $   (downward)

Magnetic field at point $P$ due to the wire having current $i$ is given as

$ B_i=\frac{\mu_0}{2 \pi} \cdot \frac{i}{y} $     (upward)

According to question, net magnetic field at point $P$ is zero, hence

$ \begin{array}{rlrl} & B_I =B_i \\ & \Rightarrow \frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y} =\frac{\mu_0}{2 \pi} \cdot \frac{i}{y} \\ & \Rightarrow \frac{I}{x+y} =\frac{i}{y} \Rightarrow i=I\left(\frac{y}{x+y}\right) \end{array} $