Moving Charges and Magnetism

Given, Velocity of proton, $\mathbf{v}=2 \hat{\mathbf{i}}$

Magnetic field $\mathbf{B}=(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \mu \mathrm{T}$

Electric field, $\mathbf{E}=10 \hat{\mathbf{i}} \mu \mathrm{V} / \mathrm{m}$

Applied Lorentz force on the proton,

$\begin{aligned} \mathbf{F} & =q \mathbf{E}+q(\mathbf{v} \times \mathbf{B})=q[\mathbf{E}+(\mathbf{v} \times \mathbf{B})] \\ & =1.6 \times 10^{-19}[10 \times 10^{-6} \hat{\mathbf{i}}+2 \hat{\mathbf{i}} \times(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times 10^{-6}] \\ & =1.6 \times 10^{-19} \times 10^{-6}[10 \hat{\mathbf{i}}+6 \hat{\mathbf{k}}-8 \hat{\mathbf{j}}] \\ \mathbf{F} & =1.6 \times 10^{-25}[10 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}] \mathrm{N} \end{aligned}$

$\therefore$ Acceleration of proton,

$\mathbf{a}=\frac{\mathbf{F}}{m_p} \quad\left(\because m_p=1.6 \times 10^{-27} \mathrm{~kg}\right)$

$\begin{aligned} & =\frac{1.6 \times 10^{-25}[10 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}]}{1.6 \times 10^{-27}} \\ \mathbf{a} & =100[10 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}] \\ \therefore \quad a & =|\mathbf{a}|=100 \sqrt{10^2+8^2+6^2} \\ & =100 \times 14.14=1414 \mathrm{~m} / \mathrm{s}^2 \\ & \simeq 1400 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

Resultant magnetic field due to current carrying parallel wire 1 ( 10 A) and wire 2 (IA) at point $A$ is zero.

$\begin{aligned} \text{i.e.,} \quad B_1 & =B_2 \\\\ \frac{\mu_0 I_1}{2 r_1} & =\frac{\mu_0 I_2}{2 r_2} \Rightarrow \frac{I_1}{r_1}=\frac{I_2}{r_2} \\\\ \frac{10}{9 \times 10^{-2}} & =\frac{I}{27 \times 10^{-2}} \Rightarrow I=30 \mathrm{~A} \end{aligned}$

Given :

Radius of the loops: $R = 10 \text{ cm} = 0.1 \text{ m}$

Current: $I = \frac{7}{2} \text{ A}$

Distance from each loop to point $P$: $x = 5 \text{ cm} = 0.05 \text{ m}$

From the figure and using Maxwell's right-hand rule, the magnetic fields due to both coils at point $P$ will be in the same direction.

The magnetic field due to one circular loop at a point along its axis is given by :

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

Since both loops contribute equally to the magnetic field at point $P$ :

$ \text{Total magnetic field, } B = 2B_1 = 2 \left( \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \right) = \frac{\mu_0 I R^2}{(R^2 + x^2)^{3/2}} $

Substitute the values into the formula :

$ B = \frac{\mu_0 \times \frac{7}{2} \times (0.1)^2}{(0.1^2 + 0.05^2)^{3/2}} $

Simplify the denominator :

$ B = \frac{\mu_0 \times \frac{7}{2} \times 0.01}{(0.01 + 0.0025)^{3/2}} $

$ B = \frac{\mu_0 \times \frac{7}{2} \times 0.01}{(0.0125)^{3/2}} $

$ B = \frac{56 \mu_0}{\sqrt{5}} \text{ T} $

Therefore, the net magnetic field at point $P$ is :

$ \boxed{\frac{56 \mu_0}{\sqrt{5}} \text{ T}} $

Given, speed of proton, $v=4.5 \times 10^5 \mathrm{~m} / \mathrm{s}$

Electrostatic force between two protons,

$F_e=k \frac{e^2}{r^2}$ ..... (i)

Magnetic field produced due to moving proton of speed $v$,

$B=\frac{\mu_0}{4 \pi} \cdot \frac{e v}{r^2}$ ..... (ii)

$\therefore$ Magnetic force on the proton,

$\begin{aligned} F_m & =B e v \\ & =\frac{\mu_0}{4 \pi} \cdot \frac{e v}{r^2} \cdot e v \quad \text{[From Eq. (ii)]} \\ F_m & =\frac{\mu_0}{4 \pi} \cdot \frac{e^2 v^2}{r^2} \quad \text{.... (iii)} \end{aligned}$

From Eqs. (i) and (iii), we get

$\begin{aligned} & \frac{F_e}{F_m}=\frac{\frac{k e^2}{r^2}}{\frac{\mu_0}{4 \pi} \cdot \frac{e^2 v^2}{r^2}}=\frac{k \cdot 4 \pi}{v^2 \cdot \mu_0} \\ & \frac{F_e}{F_m}=\frac{9 \times 10^9 \times 4 \times 3.14}{\left(4.5 \times 10^5\right)^2 \times 4 \pi \times 10^{-7}}=4.4 \times 10^5 \end{aligned}$

When an electron enters into a uniform magnetic field in perpendicular direction to the direction of magnetic field, then a magnetic force acts on the electron in perpendicular direction of both direction of magnetic field and direction of velocity of electron and direction of force can be determine by Fleming’s left hand rule. We know that, when direction of force on a particle is in perpendicular to the direction of velocity, then particle moves in uniform circular motion. Therefore, electron moving perpendicular to magnetic field (B) will perform circular motion. Hence, both assertion and reason are true and reason is correct explanation of assertion.

When a charge particle of charge $(q)$ is released from rest $(u=0)$ in the uniform magnetic field $(B)$, then magnetic force on the charged particle. $F=B q u \sin \theta=0$, because $u=0$. Therefore, the charge particle will not move on circular path. In the magnetic field, force on charge particle always acts in perpendicular direction to the direction of velocity of charge particle, therefore work done by magnetic field on charge particle is zero. Hence, assertion and reason both are false.

$\therefore B(\text { at } P)=\frac{\mu_0 I}{4 \pi d}\left(\cos \theta_1-\cos \theta_2\right)$

In given case,

$ \begin{gathered} d=R \sin 45^{\circ}=\frac{R}{\sqrt{2}} \\\\ \theta_1=135^{\circ}, \theta_2=180^{\circ} \end{gathered}$

$\begin{aligned} \therefore B(\text { at } P) & =\frac{\mu_0 I}{4 \pi\left(\frac{R}{\sqrt{2}}\right)}\left[\cos 135^{\circ}-\cos 180^{\circ}\right] \\\\ & =\frac{\mu_0 I}{4 \pi R} \sqrt{2}\left(\frac{-1}{\sqrt{2}}-(-1)\right) \\\\ & =\frac{\mu_0 I}{4 \pi R} \sqrt{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\\\ \text { or } B(\text { at } P) & =\frac{\mu_0 I}{4 \pi R}(\sqrt{2}-1) \mathrm{T} \end{aligned}$

Here, $d l=d x=1 \mathrm{~cm}=10^{-2} \mathrm{~m}$,

$i=10 \mathrm{~A}, r=0.5 \mathrm{~m}$

$\begin{aligned} \therefore \quad d B & =\frac{\mu_0}{4 \pi} \cdot \frac{i(d l \times r)}{r^3} \\ & =\frac{\mu_0}{4 \pi} \cdot \frac{i d l}{r^2}(\hat{\mathbf{i}} \times \hat{\mathbf{j}})=\frac{\mu_0}{4 \pi} \cdot \frac{i d l}{r^2} \hat{\mathbf{k}} \\ & =\frac{10^{-7} \times 10 \times 10^{-2} \sin 90^{\circ}}{(0.5)^2} \hat{\mathbf{k}} \\ & =4 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T} \end{aligned}$

$\therefore$ Magnetic flux, $\phi=B A \cos \theta$

For the top surface, the angle between normal to the surface and the $X$-axis is $\theta=60^{\circ}$.

$\begin{aligned} \therefore \phi & =0.2 \times\left(10 \times 10 \times 10^{-4}\right) \times \cos 60^{\circ} \\\\ & =10^{-3} \mathrm{~Wb}=1 \mathrm{~m}-\mathrm{Wb} \end{aligned}$

A moving charge is equivalent to a current.

$\therefore$ It produce a field and can interacts with external magnetic field.

According to Biot-Savart Law, the magnetic induction at a point to a current carrying element $i \delta l$ is given by

$\delta \mathbf{B}=\frac{\mu_0}{4 \pi} \frac{i \delta \mathbf{l} \times \mathbf{r}}{r^3} \text { or } B=\frac{\mu_0}{4 \pi} \frac{i \delta l \sin \theta}{r^2}$

Directed normal to plane containing $\delta \mathbf{l}$ and $\mathbf{r} \theta$ being angle between $\delta \mathbf{l}$ and $\mathbf{r}$

Magnetic field due to linear portions of wire If we consider any element of length $\delta \mathbf{l}$ of either linear portion $a b$ or $d e$, the angle between $\delta \mathbf{l}$ and $\mathbf{r}$ is 0 or $\pi$ therefore,

$\sin \theta=0$

Hence, magnetic induction due to both linear portions of wire is

$\frac{\mu_0}{4 \pi} \frac{i \delta l \sin \theta}{r}=0$

Field due to semi-circular are

Now angle between a current element $\delta \mathbf{l}$ of semi-circular arc and the radius vector of the element to point $c$ is $\pi / 2$. Therefore, the magnitude of magnetic induction $B$ at $O$ due to this element is

$\delta B=\frac{\mu_0}{4 \pi} \frac{i \delta l \sin \pi / 2}{r^2}=\frac{\mu_0 i \delta l}{4 \pi r^2}$

Hence, magnetic induction due to whole semi-circular loop is

$\begin{aligned} B & =\Sigma \delta B=\Sigma \frac{\mu_0}{4 \pi} \frac{i \delta l}{r^2} \\ & =\frac{\mu_0}{4 \pi} \frac{i}{r^2} \Sigma \delta l \\ & =\frac{\mu_0 i}{4 \pi r^2}(\pi r)=\frac{\mu_0 i}{4 r} \end{aligned}$

Force exerted on current carrying conductor $(F)=B I l$

$\begin{gathered} \text { Given, } B=3 \times 10^4 e^{0.2 x} \text { ay } \\\\ I=10.0 \mathrm{~A} \end{gathered}$

$\text { Average power }=\frac{\text { Work done }}{\text { Time taken }}$

$\begin{aligned} P & =\frac{1}{t} \int_\limits0^2 F \cdot d x=\frac{1}{t} \int_\limits0^2 B(x) I L \cdot d x \\\\ P & =\frac{1}{5 \times 10^{-3}} \int_\limits0^2 3 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x \\\\ & =9\left[1-e^{-0.4}\right] \\\\ & =9\left[1-\frac{1}{e^{0.4}}\right]=2.967 \mathrm{~W}=2.97 \mathrm{~W} \end{aligned}$

Cyclotron is not suitable for accelerating electrons. The reason is that due to small mass, the speed of electrons increases rapidly. Likewise due to the quick relativistic variation in their mass, the electron get out of step with the oscillating electric field.