Magnetic Effect of Current

Consider a long, straight wire of radius R carrying a steady current I. Since the current distribution is uniform, the current density (J) is constant throughout the cross-section.

The current density J is given by :

$ \mathrm{J}=\frac{\mathrm{I}}{\text { Total Area }}=\frac{\mathrm{I}}{\pi \mathrm{R}^2} $

Ampère's Law states that the line integral of the magnetic field B around a closed loop (Amperian loop) is equal to $\mu_0$ times the current enclosed ( $\mathrm{I}_{\mathrm{enc}}$ ) by that loop :

$ \oint \mathrm{B} \cdot \mathrm{dl}=\mu_0 \mathrm{I}_{\mathrm{enc}} $

To find the field inside, consider a circular Amperian loop of radius $r$ (where $r < R$ ) centered on the wire's axis.

$ \oint \mathrm{B} \mathrm{dl}=\mathrm{B} \oint \mathrm{dl}=\mathrm{B}(2 \pi \mathrm{r}) $

The current enclosed is,

$ I_{e n c}=J \times(\text { Area of loop })=\left(\frac{I}{\pi R^2}\right) \times\left(\pi r^2\right) $

$\Rightarrow $ $ I_{\mathrm{enc}}=\frac{I r^2}{R^2} $

Using Ampère's Law :

$B(2 \pi r)=\mu_0\left(\frac{I r^2}{R^2}\right)$

$\Rightarrow $ $\mathrm{B}=\frac{\mu_0 \mathrm{Ir}^2}{2 \pi \mathrm{r}^2}$

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

The magnetic field inside the wire increases linearly with the distance r from the center.

$ B \propto r $

At the axis of the conductor ( $r=0$ ), the magnetic field $B$ becomes zero (minimum). Hence, statement $B$ is incorrect and statement D is correct.

For a point outside the conductor $(\mathrm{r}>\mathrm{R})$ :

Since the Amperian loop (radius $r$ ) completely surrounds the entire cross-section of the cylinder (radius R), it encloses the total current I flowing through the conductor.

$ \mathrm{I}_{\mathrm{enc}}=\mathrm{I} $

Using Ampère's Law equation :

$ \mathrm{B}(2 \pi \mathrm{r})=\mu_0 \mathrm{I} $

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

This shows that B is inversely proportional to $\mathrm{r}\left(\mathrm{B} \propto \frac{1}{\mathrm{r}}\right)$.

The field is maximum at the surface ( $\mathrm{r}=\mathrm{R}$ ) and decreases when moving further away. Hence, statement C is incorrect.

The field distribution depends on the distance from the axis, not the length/midpoint of the conductor. Hence, statement (A) is incorrect.

Since, the field varies linearly inside and inversely outside; it is not uniform. Hence, option E is incorrect.

Therefore, only statement D is correct.

Hence, the correct option is (D).

The magnetic field at the centre of a circular loop of radius R carrying current i is given by:

$ \mathrm{B}_{\mathrm{c}}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} $

We are given that $\mathrm{B}_{\mathrm{c}}=16 \mu \mathrm{~T}$.

The magnetic field at a point on the axis at a distance x from the centre is given by:

$ \mathrm{B}_{\mathrm{a}}=\frac{\mu_0 \mathrm{i} \mathrm{R}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{\frac{3}{2}}} $

The ratio of magnetic field at the centre to the axis,

$\frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}}=\frac{\left(\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}}\right)}{\left(\frac{\mu_0 \mathrm{i} \mathrm{R}^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{\frac{3}{2}}}\right)}$

$\Rightarrow \mathrm{B}_{\mathrm{a}}=\mathrm{B}_{\mathrm{c}}\left[\frac{\mathrm{R}^3}{\left(\mathrm{R}^2+\mathrm{x}^2\right)^{3 / 2}}\right]$

Given the distance $x=\sqrt{3} R$,

$ \Rightarrow B_a=B_c\left[\frac{R^3}{\left(R^2+(\sqrt{3} R)^2\right)^{3 / 2}}\right]=\frac{B_c}{8} $

$\Rightarrow $ $ \mathrm{B}_{\mathrm{a}}=\frac{16 \mu \mathrm{~T}}{8}=2 \mu \mathrm{~T} $

Hence, the correct option is (B).

Using the formula for the force per unit length between two parallel current-carrying wires :

$ \frac{F}{l}=\frac{\mu_0 I_1 I_2}{2 \pi r} $

Wires P and Q carry currents in opposite directions (3 A up and 1 A down). Parallel wires with opposite currents repel each other. Therefore, wire $P$ pushes wire $Q$ to the right (towards wire $R$ ) with a force $F_{P Q}$

Given $\mathrm{I}_{\mathrm{P}}=3 \mathrm{~A}, \mathrm{I}_{\mathrm{Q}}=1 \mathrm{~A}, \mathrm{r}_{\mathrm{PQ}}=3 \mathrm{~cm}=0.03 \mathrm{~m}$

So, the magnitude of force on 15 cm of wire Q due to wire P is,

$ \mathrm{F}_{\mathrm{PQ}}=\frac{\left(4 \pi \times 10^{-7}\right) \times 3 \times 1 \times 0.15}{2 \pi \times 0.03} \mathrm{~N}=3 \times 10^{-6} \mathrm{~N} $

Wires Q and R carry currents in the same direction ( 1 A down and 2 A down). Parallel wires with samedirection currents attract each other. Therefore, wire $R$ pulls wire $Q$ to the right (towards wire $R$ ) with force $F_{R Q}$.

Given that $\mathrm{I}_{\mathrm{R}}=2 \mathrm{~A}, \mathrm{I}_{\mathrm{Q}}=1 \mathrm{~A}, \mathrm{r}_{\mathrm{RQ}}=2 \mathrm{~cm}=0.02 \mathrm{~m}$

$ \mathrm{F}_{\mathrm{RQ}}=\frac{\left(4 \pi \times 10^{-7}\right) \times 2 \times 1 \times 0.15}{2 \pi \times 0.02} $

$\Rightarrow $ $ \mathrm{F}_{\mathrm{RQ}}=\frac{2 \times 10^{-7} \times 0.30}{0.02}=2 \times 10^{-7} \times 15=3 \times 10^{-6} \mathrm{~N} $

Since both forces acting on wire $Q$ are directed towards the right, the net force will be towards $R$.

$ \mathrm{F}_{\mathrm{net}}=\mathrm{F}_{\mathrm{PQ}}+\mathrm{F}_{\mathrm{RQ}} $

$\Rightarrow $ $\mathrm{F}_{\text {net }}=3 \times 10^{-6}+3 \times 10^{-6}$

$ F_{n e t}=6 \times 10^{-6} \mathrm{~N} $

Hence, the net force acting on wire $Q$ is $6 \times 10^{-6} \mathrm{~N}$ towards $R$.

Hence, the correct option is (B).

The magnetic field (B) on the axis of a circular loop of radius $r$ at a distance $x$ from its center is given by :

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

The point O is at a distance $\mathrm{x}=\mathrm{r}$ from the center of both loops P and Q .

For loop P :

Current is $\mathrm{i}=\mathrm{I}$.

The magnetic field $\mathrm{B}_{\mathrm{P}}$ at O due to loop P is :

$ \mathrm{B}_{\mathrm{P}}=\frac{\mu_0 \mathrm{Ir}^2}{2\left(\mathrm{r}^2+\mathrm{r}^2\right)^{3 / 2}}=\frac{\mu_0 \mathrm{I}}{4 \sqrt{2} \mathrm{r}} $

Using the right-hand thumb rule, the current is clockwise as seen from O , so the field $\mathrm{B}_{\mathrm{P}}$ due to loop P points towards P.

For loop Q :

Current is $\mathrm{i}=4 \mathrm{I}$.

The magnetic field $\mathrm{B}_{\mathrm{Q}}$ at O due to loop Q is:

$ B_Q=\frac{\mu_0(4 I) r^2}{2\left(r^2+r^2\right)^{3 / 2}}=\frac{4 \mu_0 I}{4 \sqrt{2} r} $

Using the right-hand thumb rule, the current is clockwise as seen from O , so the field $\mathrm{B}_{\mathrm{Q}}$ due to loop Q points towards Q.

The net magnetic field $\left(B_{n e t}\right)$ at $O$ is the vector sum of these opposing fields. Since $B_Q>B_P$, the net field will be in the direction of $\mathrm{B}_{\mathrm{Q}}$ (towards Q ):

$ B_{n e t}=B_Q-B_P=\frac{4 \mu_0 I}{4 \sqrt{2} r}-\frac{\mu_0 I}{4 \sqrt{2} r}=\frac{3 \mu_0 I}{4 \sqrt{2} r} $

Therefore, the resulting magnetic field at the midpoint is $\frac{3 \mu_0 I}{4 \sqrt{2} r}$ towards $Q$.

Hence, the correct option is (B).

Let L be the side of the equilateral triangle, where $\mathrm{L}=4 \sqrt{3} \mathrm{~cm}=4 \sqrt{3} \times 10^{-2} \mathrm{~m}$.

The centroid $O$ is equidistant from all three sides. Let $r$ be the perpendicular distance from the centroid to any side.

In an equilateral triangle :

$ \tan 60^{\circ}=\sqrt{3}=\frac{L / 2}{r} \Rightarrow r=\frac{L}{2 \sqrt{3}} $

Substituting the value of L :

$ \mathrm{r}=\frac{4 \sqrt{3} \times 10^{-2}}{2 \sqrt{3}}=2 \times 10^{-2} \mathrm{~m} $

The magnetic field $B_1$ due to a straight wire of finite length at a distance $r$ is given by :

$ \mathrm{B}_1=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{r}}\left(\sin \theta_1+\sin \theta_2\right) $

Here, $\theta_1=\theta_2=60^{\circ}$

$ B_1=\frac{\mu_0 I}{4 \pi r}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)=\frac{\mu_0 I}{4 \pi r}\left(2 \sin 60^{\circ}\right) $

$\Rightarrow $ $B_1=\frac{\mu_0 I}{4 \pi r}\left(2 \times \frac{\sqrt{3}}{2}\right)=\frac{\mu_0 I \sqrt{3}}{4 \pi r}$

Since there are three sides and the current flows in a way that all three fields at the centroid point in the same direction,

$ B_{n e t}=3 \times B_1=\frac{3 \sqrt{3} \mu_0 I}{4 \pi r} $

Substituting the given values :

• 1. $\mu_0=4 \pi \times 10^{-7}$

• 2. $\mathrm{I}=2 \mathrm{~A}$

• 3. $\mathrm{r}=2 \times 10^{-2} \mathrm{~m}$

$\mathrm{B}_{\mathrm{net}}=\frac{3 \sqrt{3} \times\left(4 \pi \times 10^{-7}\right) \times 2}{4 \pi \times\left(2 \times 10^{-2}\right)} \mathrm{T}$

$\Rightarrow $ $\mathrm{B}_{\mathrm{net}}=\frac{3 \sqrt{3} \times 10^{-7} \times 2}{2 \times 10^{-2}} \mathrm{~T}$

$\Rightarrow $ $\mathrm{B}_{\mathrm{net}}=3 \sqrt{3} \times 10^{-5} \mathrm{~T}=\alpha \times 10^{-5} \mathrm{~T} $

$\Rightarrow \alpha=3 \sqrt{3}$

The segments AB and DE together form an infinitely long straight wire.

For an infinite wire the magnitude of the magnetic field at distance r is given as

$ \mathrm{B}_{\text {straight }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} $

Using the Right-Hand Thumb Rule, if the thumb points in the direction of current I (along +y ), the fingers curl out of the page at point O , i.e., in the -x direction.

$ \overrightarrow{\mathrm{B}}_{\text {straight }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \hat{\imath} $

The circular part is a nearly complete loop of radius r .

For a circular loop at its centre the magnitude of the magnetic field is given as,

$ \mathrm{B}_{\text {loop }}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}} $

Using the Right-Hand Rule for loops, curling the fingers in the direction of the current (clockwise in the yz -plane) the thumb points into the page, which is the -x direction.

$ \overrightarrow{\mathrm{B}}_{\mathrm{loop}}=-\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}} \hat{\mathrm{i}} $

The total field is the vector sum of both parts :

$ \overrightarrow{\mathrm{B}}_{\text {total }}=\overrightarrow{\mathrm{B}}_{\text {straight }}+\overrightarrow{\mathrm{B}}_{\mathrm{loop}} $

$\Rightarrow $ $\overrightarrow{\mathrm{B}}_{\mathrm{total}}=\left(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}\right) \hat{\mathrm{i}}+\left(-\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\right) \hat{\mathrm{i}}$

$\Rightarrow $ $\overrightarrow{\mathrm{B}}_{\text {total }}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left[\frac{1}{\pi}-1\right] \hat{\mathrm{i}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}[1-\pi] \hat{\mathrm{i}}=-\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}[\pi-1] \hat{\mathrm{i}}$

So, the magnetic field at the centre O of the circular loop is $=-\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}[\pi-1] \hat{1}$.

Hence, the correct option is (B).

A current-carrying solenoid produces a magnetic field ( $\overrightarrow{\mathrm{B}}$ ). For an ideal solenoid, the magnetic field inside is uniform and directed parallel to its axis.

The magnetic force $\left(\vec{F}_m\right)$ on a moving charge $Q$ with velocity $\vec{v}$ is given by the Lorentz force formula :

$ \overrightarrow{\mathrm{F}}_{\mathrm{m}}=\mathrm{Q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) $

In this case, the charged particle is released from rest along the axis. The magnetic field $\vec{B}$ is also along the axis. Therefore, the velocity $\vec{v}$ (once the particle starts moving) and the magnetic field $\vec{B}$ are parallel to each other.

The magnitude of the magnetic force is the cross product of $\mathrm{Q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$,

$ F_m=Q v B \sin \theta $

Since the angle $\theta$ between the velocity and the magnetic field is $0^{\circ}$ :

$ \mathrm{F}_{\mathrm{m}}=\mathrm{Q} v \mathrm{~B} \sin 0^{\circ} $

The magnetic field exerts no force on the particle moving along the axis.

The solenoid is placed vertically. When the particle of mass $m$ is released, the only non-zero force acting on it is the force of gravity $\left(\overrightarrow{\mathrm{F}}_{\mathrm{g}}\right)$ :

$ \overrightarrow{\mathrm{F}}_{\mathrm{g}}=\mathrm{mg} $

This force acts vertically downwards along the axis of the solenoid.

According to Newton's Second Law ( $\mathrm{F}_{\text {net }}=\mathrm{ma}$ ) :

$ \mathrm{mg}=\mathrm{ma} $

$ \Rightarrow \mathrm{a}=\mathrm{g} $

Since the magnetic force is zero the particle falls freely under gravity. Thus, its acceleration is exactly equal to g.

Hence, the correct option is (D).

When a current-carrying loop is placed in an external magnetic field, it experiences a torque $(\vec{\tau})$. The magnetic properties of the loop are represented by its magnetic dipole moment $(\overrightarrow{\mathrm{m}})$, defined as :

$ \overrightarrow{\mathrm{m}}=\mathrm{I} \overrightarrow{\mathrm{~A}} $

Where I is the current and $\vec{A}$ is the area vector of the loop. The direction of the area vector is perpendicular to the plane of the loop, given by the unit normal $\hat{\mathrm{n}}$.

$ \overrightarrow{\mathrm{A}}=\mathrm{A} \hat{\mathrm{n}} $

The torque is given by the cross product of the magnetic moment and the magnetic field :

$ \vec{\tau}=\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{B}} $

$\Rightarrow $ $ \vec{\tau}=(\mathrm{IA} \hat{\mathrm{n}}) \times \overrightarrow{\mathrm{B}} $

The radius of circular loop is $\mathrm{r}=2 \mathrm{~cm}=0.02 \mathrm{~m}$.

So, the area is

$ \mathrm{A}=\pi \mathrm{r}^2=\pi(0.02)^2=4 \pi \times 10^{-4} \mathrm{~m}^2 $

The unit normal is $\hat{\mathrm{n}}=\frac{\hat{\mathrm{k}}+\hat{\mathrm{t}}}{\sqrt{2}}$

And the current in loop is $\mathrm{I}=100 \sqrt{2} \mathrm{~A}$

So, the magnetic moment vector is;

$ \overrightarrow{\mathrm{m}}=(100 \sqrt{2}) \times\left(4 \pi \times 10^{-4}\right) \times\left(\frac{\hat{\imath}+\hat{\mathrm{k}}}{\sqrt{2}}\right) $

$\Rightarrow $ $ \overrightarrow{\mathrm{m}}=400 \pi \times 10^{-4}(\hat{\imath}+\hat{\mathrm{k}})=0.04 \pi(\hat{\imath}+\hat{\mathrm{k}}) \mathrm{A} \cdot \mathrm{~m}^2 $

The magnetic field in the region is $\vec{B}=B_o(3 \hat{\imath}+2 \hat{k})$ with $B_o=4 \times 10^{-3} \mathrm{~T}$

$ \overrightarrow{\mathrm{B}}=4 \times 10^{-3}(3 \hat{\imath}+2 \hat{\mathrm{k}})=(12 \hat{\imath}+8 \hat{\mathrm{k}}) \times 10^{-3} \mathrm{~T} $

Using the formula for magnetic torque,

$ \vec{\tau}=\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{B}} $

$\Rightarrow $ $\vec{\tau}=[0.04 \pi(\hat{\imath}+\hat{k})] \times\left[(12 \hat{\imath}+8 \hat{k}) \times 10^{-3}\right]$

$\Rightarrow $ $\vec{\tau}=\left(0.04 \pi \times 10^{-3}\right)[(\hat{\imath}+\hat{k}) \times(12 \hat{\imath}+8 \hat{k})]$

$\Rightarrow $ $\vec{\tau}=\left(0.04 \pi \times 10^{-3}\right)(4 \hat{\jmath})$

$\Rightarrow $ $\vec{\tau}=0.16 \pi \times 10^{-3} \hat{\jmath}$

$\Rightarrow $ $\vec{\tau}=0.16 \times 3.14 \times 10^{-3} \hat{\jmath}=0.5024 \times 10^{-3} \hat{\jmath}$

$\Rightarrow $ $ \vec{\tau}=5024 \times 10^{-7} \hat{\jmath} $

Therefore, the torque experienced by the loop is $5024 \times 10^{-7} \hat{\mathrm{j}}$.

Hence, the correct option is (D).

According to the Ampere's circuital Law, the magnetic field B at a distance r from an infinitely long straight wire carrying current I is :

$ \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{~d} \overrightarrow{\mathrm{l}}=\mu_0 \mathrm{I} $

For a circular path of radius $r$ :

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

Let the wires be $\mathrm{W}_1$ and $\mathrm{W}_2$ separated by $\mathrm{d}=8 \mathrm{~cm}$. The midpoint is at $\mathrm{r}=4 \mathrm{~cm}=0.04 \mathrm{~m}$ from each wire.

Wire 1 carries current along out of the page. Using the right-hand rule, the field $\mathrm{B}_1$ at the midpoint points along $y$-axis.

Wire 2 carries current along into the page. Using the right-hand rule, the field $\mathrm{B}_2$ at the midpoint also points along y -axis.

Since both fields point in the same direction, the net magnetic field $\mathrm{B}_{\text {net }}=\mathrm{B}_1+\mathrm{B}_2$.

The current in the wires are given as $\mathrm{I}_1=\mathrm{I}_2=30 \mathrm{~A}$ and $\mathrm{r}=0.04 \mathrm{~m}$ :

$ B_1=B_2=\frac{\mu_0 I}{2 \pi r}=2 \times\left(\frac{\mu_0}{4 \pi}\right) \frac{I}{r} $

$\Rightarrow $ $\mathrm{B}_1=\mathrm{B}_2=2 \times 10^{-7} \times \frac{30}{0.04}=2 \times 10^{-7} \times 750=15 \times 10^{-5} \mathrm{~T}$

$\Rightarrow $ $B_{\text {net }}=15 \times 10^{-5}+15 \times 10^{-5}=30 \times 10^{-5} \mathrm{~T}$

$\Rightarrow $ $ \mathrm{B}_{\mathrm{net}}=300 \times 10^{-6} \mathrm{~T}=300 \mu \mathrm{~T} $

Therefore, the magnitude of magnetic field at the midpoint is $300 \mu \mathrm{~T}$.

Thus, the correct option is (B).

The magnetic field produced at the center of a circular wire of radius $r$ carrying current $I$ is:

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

The electric current in the circular loop is $\mathrm{I}=2 \mathrm{~A}$, the radius of the loop is $\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}$.

$ \mathrm{B}=\frac{4 \pi \times 10^{-7} \times 2}{2 \times 0.1}=4 \pi \times 10^{-6} \mathrm{~T} $

Magnetic energy is stored in the space where a magnetic field exists. The energy per unit volume is:

$ u_B=\frac{B^2}{2 \mu_0} $

For a very small volume V where the field B is approximately uniform:

$ \mathrm{U}=\mathrm{u}_{\mathrm{B}} \times \mathrm{V}=\frac{\mathrm{B}^2}{2 \mu_0} \times \mathrm{V} $

The volume of the cube is,

$ V=(1 \mathrm{~mm})^3=\left(10^{-3} \mathrm{~m}\right)^3=10^{-9} \mathrm{~m}^3 $

So, the total energy stored in the cube is,

$ U=\frac{\left(4 \pi \times 10^{-6}\right)^2}{2 \times 4 \pi \times 10^{-7}} \times 10^{-9} $

$\Rightarrow $ $U=\frac{16 \pi^2 \times 10^{-12}}{8 \pi \times 10^{-7}} \times 10^{-9}=2 \pi \times 10^{-5} \times 10^{-9}$

$\Rightarrow $ $\mathrm{U}=2 \times 3.14 \times 10^{-14}=6.28 \times 10^{-14} \mathrm{~J}=\alpha \times 10^{-14} \mathrm{~J}$

$ \Rightarrow \alpha=6.28 $

Therefore, the correct option is (A).

The change in kinetic energy can be found using the work-energy theorem.

For a charged particle in electric and magnetic fields,

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

Now, the magnetic force is always perpendicular to velocity, so it does no work.

Hence, change in kinetic energy is only due to the electric field:

$ \Delta K = q \int \vec{E}\cdot d\vec{r} $

Given,

$ \vec{E} = E_0 e^{-\lambda x}\hat{x} $

and

$ d\vec{r} = dx\,\hat{x} + dy\,\hat{y} + dz\,\hat{z} $

So,

$ \vec{E}\cdot d\vec{r} = E_0 e^{-\lambda x} dx $

Therefore,

$ \Delta K = q \int_0^L E_0 e^{-\lambda x}\,dx $

Now integrate:

$ \Delta K = qE_0 \left[\frac{1 - e^{-\lambda L}}{\lambda}\right] $

So,

$ \boxed{\Delta K = \frac{qE_0}{\lambda}\left(1-e^{-\lambda L}\right)} $

Hence the correct option is:

$ \boxed{\text{Option A}} $

Given:

• Charge, $q = 10^{-9}\,\text{C}$

• Electric field, $\vec E = 0.4\,\hat j\ \text{N/C}$

• Magnetic field, $\vec B = 4\times 10^{-3}\,\hat k\ \text{T}$

• Force, $\vec F = (4\hat i + 2\hat j)\times 10^{-10}\ \text{N}$

We use the Lorentz force formula:

$ \vec F = q\left(\vec E + \vec v \times \vec B\right) $

Since the particle is moving in the $x$-$y$ plane, let

$ \vec v = v_x \hat i + v_y \hat j $

Now calculate $\vec v \times \vec B$:

$ \vec v \times \vec B = (v_x \hat i + v_y \hat j)\times \left(4\times 10^{-3}\hat k\right) $

Using cross products:

$ \hat i \times \hat k = -\hat j,\qquad \hat j \times \hat k = \hat i $

So,

$ \vec v \times \vec B = 4\times 10^{-3}\left(v_y \hat i - v_x \hat j\right) $

Now,

$ \vec E + \vec v \times \vec B = 0.4\hat j + 4\times 10^{-3}v_y \hat i - 4\times 10^{-3}v_x \hat j $

$ = \left(4\times 10^{-3}v_y\right)\hat i + \left(0.4 - 4\times 10^{-3}v_x\right)\hat j $

Multiplying by $q = 10^{-9}$:

$ \vec F = 10^{-9}\left[\left(4\times 10^{-3}v_y\right)\hat i + \left(0.4 - 4\times 10^{-3}v_x\right)\hat j\right] $

This is equal to the given force:

$ (4\hat i + 2\hat j)\times 10^{-10} $

Now compare components.

For $\hat i$ component:

$ 10^{-9}\left(4\times 10^{-3}v_y\right) = 4\times 10^{-10} $

$ 4\times 10^{-12}v_y = 4\times 10^{-10} $

$ v_y = 100\ \text{m/s} $

For $\hat j$ component:

$ 10^{-9}\left(0.4 - 4\times 10^{-3}v_x\right) = 2\times 10^{-10} $

$ 0.4 - 4\times 10^{-3}v_x = 0.2 $

$ 4\times 10^{-3}v_x = 0.2 $

$ v_x = 50\ \text{m/s} $

Hence,

$ \vec v = 50\hat i + 100\hat j\ \text{m/s} $

So, the correct option is:

$ \boxed{50\hat i + 100\hat j} $

Option A

$\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{BC}}+\overrightarrow{\mathrm{B}}_{\mathrm{CD}}+\overrightarrow{\mathrm{B}}_{\mathrm{DA}}$

$\begin{aligned} \overrightarrow{\mathrm{B}}= & {\left[\frac{-\mu_0(2 \mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}-\frac{\mu_0(2 \mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}\right.} \\ & \left.+\frac{\mu_0(\mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}+\frac{\mu_0(\mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}\right] \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{~B}}= & {\left[\frac{-2 \sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}+\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}\right] \hat{\mathrm{k}} } \\ \overrightarrow{\mathrm{~B}}= & \frac{-\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}} \hat{\mathrm{k}} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB} \\ & \mathrm{mv}=\mathrm{qBr} \\ & \mathrm{E}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{E}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{q}^2 \mathrm{~B}^2 \mathrm{r}^2}{\mathrm{~m}^2}\right)=\frac{\mathrm{q}^2 \mathrm{~B}^2 \mathrm{r}^2}{2 \mathrm{~m}} \\ & \mathrm{E}=\left(\frac{\mathrm{q}^2 \mathrm{~B}^2}{2 \mathrm{~m}}\right) \mathrm{r}^2 \\ & \mathrm{r}^2 \propto \mathrm{E} \end{aligned}$

As $\overrightarrow{\mathrm{v}}$ is $\perp$ to $\overrightarrow{\mathrm{B}}$, so charge particle will move in circular path, whose radius is given by

$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$

Starting point $\rightarrow \mathrm{A}$

Ending point $\rightarrow \mathrm{C}$

$\therefore$ Net displacement $=\mathrm{AC}$

$\begin{aligned} & \mathrm{AC}=\mathrm{CD}-\mathrm{AD} \\ & \mathrm{AC}=\frac{2 \mathrm{mv}}{\mathrm{qB}_1}-\frac{2 \mathrm{mv}}{\mathrm{qB}_2} \\ & \mathrm{AC}=\frac{2 \mathrm{mv}}{\mathrm{qB}_1}\left[1-\frac{\mathrm{B}_1}{\mathrm{~B}_2}\right] \end{aligned}$

The explanation involves analyzing the motion of charged particles in a magnetic field using the concept of radius of curvature. When a charged particle moves perpendicularly to a magnetic field, the radius of curvature $ r $ of its path is given by:

$ r = \frac{mv}{qB} = \frac{p}{qB} $

Where:

$ m $ is the mass of the particle.

$ v $ is the velocity.

$ q $ is the charge.

$ B $ is the magnetic field strength.

$ p $ is the momentum of the particle.

From this formula, we can see that the radius of curvature $ r $ is directly proportional to the momentum $ p $ and inversely proportional to the charge $ q $.

In the given scenario:

The Oxygen ion $\mathrm{O}^{-2}$ and the Hydrogen ion $\mathrm{H}^{+}$ both enter the magnetic field with equal momentum.

Since $\mathrm{O}^{-2}$ has a charge of $-2e$ and $\mathrm{H}^{+}$ has a charge of $+e$, and given that $r \propto \frac{1}{q}$, the path of $\mathrm{O}^{-2}$ will indeed have a smaller radius of curvature compared to that of $\mathrm{H}^{+}$.

Therefore, the Assertion $ \mathbf{A} $ is true.

The Reason $ \mathbf{R} $ states that a proton and an electron with the same linear momentum will form paths of different radii upon entering a magnetic field, with the proton having a smaller radius of curvature. This reasoning is based on the same principle, since the mass and charge of the electron and proton are different. However, this explanation does not directly justify the assertion about Oxygen and Hydrogen ions, so Reason $ \mathbf{R} $ is false in the context of explaining Assertion $ \mathbf{A} $. Thus, the correct conclusion is that $\mathbf{A}$ is true, but $\mathbf{R}$ is false.

In a moving coil galvanometer, the voltage sensitivity is given by the formula:

$ \text{Voltage sensitivity} = \frac{\theta}{V} = \frac{N \cdot A \cdot B}{c \cdot R} $

Here, $ \theta $ is the deflection angle, $ V $ is the voltage, $ N $ is the number of turns, $ A $ is the coil area, $ B $ is the magnetic field, $ c $ is the torsional constant of the spring, and $ R $ is the resistance of the coil.

For the two moving coils $ \mathrm{M}_1 $ and $ \mathrm{M}_2 $, we want to determine the ratio of their voltage sensitivities. Given that the torsional constant $ c $ is the same for both coils, we can express the ratio as:

$ \text{Ratio of voltage sensitivities} = \left(\frac{N_1 \cdot A_1 \cdot B_1}{N_2 \cdot A_2 \cdot B_2}\right) \cdot \frac{R_2}{R_1} $

Plugging in the given values:

$ N_1 = 15 $

$ A_1 = 3.6 \times 10^{-3} \, \text{m}^2 $

$ B_1 = 0.25 \, \text{T} $

$ R_1 = 5 \, \Omega $

$ N_2 = 21 $

$ A_2 = 1.8 \times 10^{-3} \, \text{m}^2 $

$ B_2 = 0.50 \, \text{T} $

$ R_2 = 7 \, \Omega $

We calculate:

$ \text{Ratio} = \left(\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5}\right) \times \frac{7}{5} = \frac{1}{1} $

Thus, the ratio of the voltage sensitivities of $ M_1 $ to $ M_2 $ is 1:1.

To determine the magnetic energy stored in a solenoid filled with a material of relative permeability 2, we start with the expression for energy density in a magnetic field:

$ \frac{U}{V} = \frac{B^2}{2 \mu_{\mathrm{r}} \mu_0} $

Given that the relative permeability $\mu_{\mathrm{r}}$ is 2, this becomes:

$ \frac{U}{V} = \frac{B^2}{2 \times 2 \mu_0} = \frac{B^2}{4 \mu_0} $

The energy $U$ stored in the solenoid can be expressed as:

$ U = \frac{B^2}{4 \mu_0} \times V $

Here, $V$ is the volume of the solenoid, calculated as $A \times \ell$ (where $A$ is the cross-sectional area and $\ell$ is the length). Therefore, substituting for $V$, we get:

$ U = \frac{B^2}{4 \mu_0} \times A \ell $

This formula gives us the magnetic energy stored in the solenoid.

$\begin{aligned} & B_1=\frac{\mu_0 i}{2 R} \quad \quad B_2=B_1 \sin ^3 \theta \\ & \therefore \frac{B_2}{B_1}=\sin ^3 \theta=\left(\frac{4}{5}\right)^3=\frac{64}{125} \end{aligned}$

Maximum possible magnetic field is at the surface

$\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}$

It can be obtained inside as well as outside the wire

For inside,

$\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \\ & \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2} \end{aligned}$

For outside

$\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \mathrm{r}=2 \mathrm{a} \end{aligned}$

Correct answer $\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$

$\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{a}} \otimes \\ & \mathrm{~B}_2=\frac{\mu_0}{4 \pi} \frac{\mathrm{i}}{\mathrm{a}}\left(\frac{3 \pi}{2}\right) \otimes \\ & \mathrm{B}_3=0 \\ & \mathrm{~B}=\frac{\mu_0}{4 \pi} \frac{\mathrm{i}}{\mathrm{a}}\left(1 \frac{3 \pi}{2}\right) \otimes \end{aligned}$

For $A \to B$

$\overrightarrow V = - {v_x}\widehat i + {v_y}\widehat j$

$\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)$

$\because$ $\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$

So, $\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ {\left[ {\left( { - {v_x}} \right)\widehat i + {v_y}\widehat j} \right] \times \left( { - \widehat k} \right)} \right]$

$ = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) - {v_y}\left( {\widehat j \times \widehat k} \right)} \right]$

$\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ { - {v_x}\widehat j - {v_y}\widehat i} \right]$

$ \Rightarrow {F_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$

$ \Rightarrow m{a_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$

$ \Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$

$ \Rightarrow {{{v_x}d{v_x}} \over {{v_y}}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{dr} \over r}$

Since, $v_x^2 + v_y^2 = v_0^2$

$ \Rightarrow {v_y} = \sqrt {v_0^2 - v_x^2} $

$ \Rightarrow \int_0^{{v_0}} {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{ - {\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} } $

Let $v_0^2 - v_x^2 = {z^2}$

$ \Rightarrow - 2{v_x}d{v_x} = 2zdz$

when ${v_x} = 0,\,z = {v_0}$

${v_x} = {v_0},\,z = 0$

$ \Rightarrow \int_{{v_0}}^0 {{{zdz} \over z} = - {{{\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} } $

$ \Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{{x'} \over a}} \right)$

$ \Rightarrow \ln \left( {{{x'} \over a}} \right) = - {{2\pi m{v_0}} \over {{\mu _0}Iq}}$

$ \Rightarrow x' = ac{{ - 2\pi m{v_0}} \over {{\mu _0}Iq}}$ .... (1)

Now, for $B \to C$

$\overrightarrow v = - {v_x}\widehat i - {v_y}\widehat j$

$\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)$

$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right) = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) + {v_y}\left( {\widehat j \times \widehat k} \right)} \right]$

$ \Rightarrow \overrightarrow F = {{{\mu _0}Iq} \over {2\pi r}}\left[ { - {v_x}\widehat j + {v_y}\widehat i} \right]$

$ \Rightarrow {F_x} = {{{\mu _0}Iq} \over {2\pi r}}{v_y}$

$ \Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{{\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$

$\int_{{v_0}}^0 {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{{\mu _0}Iq} \over {2\pi m}}\int_{x'}^x {{{dr} \over r}} } $

Let $v_0^2 - v_x^2 = {z^2}$

$ - 2{v_x}d{v_x} = 2zdz$

${v_x} = {v_0} \Rightarrow z = 0$

${v_x} = 0 \Rightarrow z = {v_0}$

$ \Rightarrow - \int_0^{{v_0}} {{{zdz} \over z} = {{{\mu _0}Iq} \over {2\pi m}}\left[ {\ln r} \right]_{x'}^x} $

$ \Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{x \over {x'}}} \right)$

$ \Rightarrow x = x'{e^{ - {{2\pi m{v_0}} \over {{\mu _0}Iq}}}}$ .... (2)

From (1) & (2), $x = a{e^{ - {{4\pi m{v_0}} \over {{\mu _0}Iq}}}}$

$\begin{aligned} & I_A=\frac{N q}{\frac{2 \pi}{\omega}} \\ & I_A=\frac{N q \omega}{2 \pi}, I_B=0 \\ & I_A-I_B=\frac{N q \omega}{2 \pi} \end{aligned}$

According to the Lorentz force law, the force experienced by a charged particle, such as an electron, moving in a magnetic field is given by the equation:

$ \overline{\mathrm{F}} = \mathrm{q}(\overline{\mathrm{v}} \times \overline{\mathrm{B}}) $

Here, $ \overline{\mathrm{F}} $ is the force exerted on the particle, $ \mathrm{q} $ is the charge of the particle, $ \overline{\mathrm{v}} $ is the velocity vector of the particle, and $ \overline{\mathrm{B}} $ is the magnetic field vector. The symbol $ \times $ indicates the cross product, meaning the force is perpendicular to both the velocity and magnetic field vectors.

Given this, for the force $ \overline{\mathrm{F}} $ to be zero:

The velocity $ \overline{\mathrm{v}} $ and the magnetic field $ \overline{\mathrm{B}} $ must be parallel (or anti-parallel). This condition will result in the angle $ \theta = 0^\circ $ or $ 180^\circ $.

Under these circumstances, the electron will not experience any magnetic force and will continue to move in a straight line with constant velocity, as the cross product becomes zero. Therefore, the reason (R) is indeed a valid explanation: the magnetic field being along the direction of the electron's velocity means no perpendicular force is applied to change its path.

$\begin{aligned} & \mathrm{B}_{\text {in }}=\frac{\mu_0(\overline{\mathrm{~J}} \times \overline{\mathrm{r}})}{2} \\ & \mathrm{~B}_{\text {in }} \propto \mathrm{r} \\ & \mathrm{~B}_{\text {out }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \mathrm{~B}_{\text {net }} \propto \frac{1}{\mathrm{r}} \end{aligned}$

When an electron is projected perpendicular to a uniform magnetic field $ B $, it travels in a circular path. According to Bohr's quantization rule, the radius of the electron's orbit in the first excited state can be determined using the following approach:

The radius $ r $ in a magnetic field is expressed as:

$ r = \frac{mv}{eB} $

where $ m $ is the mass of the electron, $ v $ is its velocity, $ e $ is its charge, and $ B $ is the magnetic field strength.

According to Bohr's quantization condition:

$ mvr = \frac{nh}{2\pi} $

Substituting the expression for $ r $:

$ (eBr)r = \frac{nh}{2\pi} $

This simplifies to the expression for the radius:

$ r = \sqrt{\frac{nh}{2\pi eB}} $

In the first excited state, the principal quantum number $ n $ is 2. So, the radius becomes:

$ r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}} $

Thus, for the first excited state, the radius of the electron's orbit is $ \sqrt{\frac{h}{\pi eB}} $.

To solve for the ratio of the radii of deutron and proton paths in a magnetic field, we use the formula for the radius $r$ of the circular path of a charged particle moving perpendicular to a uniform magnetic field:

$r = \frac{mv}{qB}$

where:

• $m$ is the mass of the particle,
• $v$ is the velocity of the particle,
• $q$ is the charge of the particle, and
• $B$ is the magnetic field strength.

The proton and the deutron are given to have the same kinetic energy. The kinetic energy $K$ of a particle is given by:

$K = \frac{1}{2}mv^2$

From the kinetic energy, we can express the velocity as:

$v = \sqrt{\frac{2K}{m}}$

Since both particles have the same kinetic energy and the charge of the deutron is the same as the charge of the proton $q = e$, but the mass of the deutron is twice that of the proton ($m_d = 2m_p$), substituting the expression for $v$ in the radius formula, we get:

For the deutron:

$r_d = \frac{m_d\sqrt{2K/m_d}}{eB} = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_d}$

For the proton ($m_p = m$):

$r_p = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_p}$

The ratio of the radius of the deutron path $r_d$ to the radius of the proton path $r_p$ is therefore:

$\frac{r_d}{r_p} = \frac{\sqrt{m_d}}{\sqrt{m_p}} = \sqrt{\frac{2m_p}{m_p}} = \sqrt{2}$

So, the correct answer is:

Option C: $\sqrt{2}: 1$.

Let's analyze each statement in detail to determine the correct answer:

Statement (I) : "When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account." This statement is true. According to physics, particularly when dealing with electromagnetism and Maxwell's equations, the momentum of the electromagnetic field plays a critical role in conserving momentum in systems where electromagnetic forces are at play. In situations where electric and magnetic fields vary with time, they can carry momentum. Thus, for the conservation laws to hold, including Newton's third law which states that for every action, there is an equal and opposite reaction, the momentum carried by the electromagnetic fields must be included. This is essential in scenarios such as radiation pressure where light (which can be considered an electromagnetic wave) exerts pressure on surfaces, thereby imparting momentum.

Statement (II) : "Ampere's circuital law does not depend on Biot-Savart's law." This statement is false. Historically and mathematically, Ampère's Circuital Law and Biot-Savart Law are closely related in the context of classical electromagnetism. Ampère's Law, particularly in its integral form, relates the integrated magnetic field around a closed loop to the electric current passing through the loop. Biot-Savart Law, on the other hand, is used to calculate the magnetic field generated by a current-carrying element at a point in space. Although Ampère's Law can be derived without directly invoking the Biot-Savart Law, the fundamental understanding and derivations of magnetic fields due to currents, as presented in many textbooks and formulations, show that both laws are manifestations of how moving charges produce magnetic fields. Moreover, the formulation of Ampère's Law was later extended by Maxwell (Maxwell's addition) to include the concept of displacement current, linking it more fundamentally to the changing electric fields and closing a conceptual loop that ties it to the broader electromagnetic theory that includes the effects described by the Biot-Savart Law.

Given the above explanations:

Option D (Statement I is true but Statement II is false) is the correct answer.

To find the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ distances from the axis of a long straight wire, we use Ampère's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.

Ampère’s Law is given by:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$

where:

$\vec{B}$ is the magnetic field,

$\mu_0$ is the permeability of free space,

$I_{\text{enc}}$ is the enclosed current.

For a point inside the wire (at radius $r=a/2$):

The current enclosed by a radius $r$ is proportional to the area of the cross-section at radius $r$.

The area of the cross-section at radius $r$ is given by:

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

The total current $I$ is uniformly distributed, thus the current enclosed $I_{\text{enc}}$ at radius $r = \frac{a}{2}$ is:

$I_{\text{enc}} = I \times \frac{\text{Area enclosed}}{\text{Total area}} = I \times \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4}$

Applying Ampère’s Law inside the conductor, we get:

$B \cdot 2 \pi \left( \frac{a}{2} \right) = \mu_0 \left( \frac{I}{4} \right)$

So,

$B \cdot \pi a = \frac{\mu_0 I}{4}$

Therefore, the magnetic field inside the wire at $r = \frac{a}{2}$ is:

$B_{\frac{a}{2}} = \frac{\mu_0 I}{4 \pi a}$

For a point outside the wire (at radius $r = 2a$):

The total current enclosed by a radius $r = 2a$ is the entire current $I$.

Applying Ampère’s Law outside the conductor, we get:

$B \cdot 2 \pi (2a) = \mu_0 I$

So,

$B \cdot 4 \pi a = \mu_0 I$

Therefore, the magnetic field outside the wire at $r = 2a$ is:

$B_{2a} = \frac{\mu_0 I}{4 \pi a}$

Hence, the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ is:

$\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1:1$

So, the correct option is:

Option C: $1:1$

$\begin{aligned} & B=\frac{u_0}{4 \pi} \frac{i d l \sin \theta}{r^2} \\ & \Rightarrow B=\frac{10^{-7} \times 10 \times 10 \quad \times 1}{\frac{1}{4}}=4 \times 10^{-8} \mathrm{~T} \end{aligned}$

The correct expressions for the electrostatic force, $\vec{F}_1$, and the magnetic force, $\vec{F}_2$, acting on a charge $q$ moving with velocity $\vec{v}$, are given by the Lorentz force law. This law states that the total force acting on a charged particle in both an electric field and a magnetic field is the sum of an electrostatic force due to the electric field and a magnetic force due to the magnetic field.

The electrostatic force is given by:

$\vec{F}_1 = q \vec{E}$

where $\vec{E}$ is the electric field.

The magnetic force is given by:

$\vec{F}_2 = q(\vec{v} \times \vec{B})$

where $\vec{B}$ is the magnetic field, and $\times$ denotes the cross product, indicating that the magnetic force is perpendicular both to the direction of the velocity of the charge and the direction of the magnetic field.

Therefore, the correct option is:

Option C: $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B})$

Options A, B, and D are incorrect because they do not accurately reflect the definitions of electrostatic and magnetic forces as described by the Lorentz force law.

Let's dig into how the magnetic field behaves in a coaxial cable in relation to the given options. The principle to consider here is Ampère's law, which states that the magnetic field in a loop surrounding a current is proportional to the amount of current enclosed. When we apply this to a coaxial cable, we must look at different regions within the cable.

Option A: Inside the outer conductor

The magnetic field within the outer conductor is not zero because the current in the outer conductor itself contributes to the magnetic field in that region. However, considering the symmetric distribution of current and the geometry of the coaxial cable, there may be a varying magnetic field within the conductor depending on the distance from the center axis.

Option B: Outside the cable

Outside the coaxial cable, the net current enclosed by a path enclosing both conductors is zero because the current in the inner conductor flows in the opposite direction to the equally magnitude current in the outer conductor. These currents being equal and opposite in direction cancel each other out, leading to a net enclosed current of zero. According to Ampère's law, if the net enclosed current is zero, the magnetic field in that space is also zero. Therefore, the magnetic field is zero outside the cable.

Option C: In between the two conductors

In the space between the two conductors, the magnetic field is not zero. This region only encloses the current from the inner conductor. The magnetic field in this region is due to the current in the inner conductor and follows the right-hand rule, which would result in concentric circles of magnetic field around the inner conductor. Since only the inner conductor's current contributes to the magnetic field in this space, Ampère's law suggests that there is a non-zero magnetic field in this region.

Option D: Inside the inner conductor

Within the inner conductor, the magnetic field is not necessarily zero. Like within the outer conductor, the magnetic field inside the inner conductor will depend on the distribution of the current within that conductor. Utilizing the formula derived from Ampère's law for a cylindrical conductor with a uniform current distribution, the magnetic field inside the conductor increases linearly with the distance from the center axis up to the conductor's surface.

In conclusion, the correct option, based on Ampère's law and the principle that the net current enclosed determines the magnetic field outside the current's path, is:

Option B: Outside the cable

This is because the equal and opposite currents in the inner and outer conductors cancel each other, leading to a net magnetic field of zero outside the coaxial cable.

For this scenario, it's important to recall how magnetic fields influence the motion of charged particles, and the configuration of the magnetic field within a solenoid. Inside a long solenoid, the magnetic field lines run parallel to the axis of the solenoid. The strength of this field is uniform and depends on the current running through the solenoid's coils and the number of turns per unit length but is independent of the position inside the solenoid as long as one is sufficiently far from the ends.

The force experienced by a charged particle moving in a magnetic field is given by the Lorentz force equation: $\vec{F} = q(\vec{v} \times \vec{B})$ where

• $q$ is the charge of the particle,


• $\vec{v}$ is the velocity of the particle, and


• $\vec{B}$ is the magnetic field.

For an electron moving along the axis inside a solenoid:

• The magnetic field ($\vec{B}$) inside the solenoid is parallel to the axis of the solenoid.


• The velocity of the electron ($\vec{v}$) is also parallel to the axis of the solenoid and hence parallel to $\vec{B}$.

Since the cross product of two parallel vectors (in this case, $\vec{v}$ and $\vec{B}$) is zero, the Lorentz force ($\vec{F} = q(\vec{v} \times \vec{B})$) acting on the electron will be zero. Therefore, the electron will not experience any force due to the magnetic field of the solenoid, as there is no component of its velocity that is perpendicular to the magnetic field within the solenoid.

Given the above explanation, the correct option is:

Option D: the electron will continue to move with uniform velocity along the axis of the solenoid.

This is because, in the absence of any force acting on it, the electron will maintain its state of motion according to Newton's first law of motion, which states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

$\begin{aligned} & \overrightarrow{\mathrm{M}}=\mathrm{i} \overrightarrow{\mathrm{A}} \\ & =5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}}) \\ & =0.1(-\hat{\mathrm{i}}) \\ & \vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}}) \\ & =2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m} \end{aligned}$

Note : Direction of magnetic field is in $+\hat{k}$

$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{i} \vec{\ell} \times \overrightarrow{\mathrm{B}} \\ & \ell=2 \mathrm{R} \\ & \overrightarrow{\mathrm{F}}=-2 \mathrm{iRB} \hat{\mathrm{j}} \end{aligned}$