Physics 4 Marks Questions

1. If the magnetic field at the centre of the coil is considered, then x = 0.

$\therefore\text{B}=\frac{\mu_{0}\text{IR}^{2}\text{N}}{2\text{R}^{3}}=\frac{\mu_{0}\text{IN}}{2\text{R}}$

This is the familiar result for magnetic field at the centre of the coil.

2. Radis of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both coils = I

Distance between both the coils = R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of $\frac{\text{R}}{2}+\text{d}$  from point Q.

$\therefore$ Magnetic field at point Q is given as:

$\text{B}_{1}=\frac{\mu_{0}\text{NIR}^{2}}{2\Big[\Big(\frac{\text{R}}{2}+\text{d}\big)^{2}+\text{R}^{2}\Big]^{\frac{3}{2}}}$

Also, the other coil is at a distance of $\frac{\text{R}}{2}-\text{d}$ from point Q.

$\therefore$ Magnetic field due to this coil is given as:

$\text{B}_{2}=\frac{\mu_{0}\text{NIR}^{2}}{2\Big[\Big(\frac{\text{R}}{2}-\text{d}\big)^{2}+\text{R}^{2}\Big]^{\frac{3}{2}}}$

Total magnetic field, $\text{B}=\text{B}_{1}+\text{B}_{2}$

$=\frac{\mu_{0}\text{IR}^{2}}{2}\bigg[\Big\{\Big(\frac{\text{R}}{2}-\text{d}\Big)^{2}+\text{R}^{2}\Big\}^{-\frac{3}{2}}+\Big\{\Big(\frac{\text{R}}{2}+\text{d}\Big)^{2}+\text{R}^{2}\Big\}^{-\frac{3}{2}}\Bigg]$

$=\frac{\mu_{0}\text{IR}^{2}}{2}\times\Big(\frac{5\text{R}^{2}}{4}\Big)^{-\frac{3}{2}}\bigg[\Big(1+\frac{4}{5}\frac{\text{d}^{2}}{\text{R}^{2}}-\frac{4}{5}\frac{\text{d}}{\text{R}}\Big)^{-\frac{3}{2}}+\Big(1+\frac{4}{5}\frac{\text{d}^{2}}{\text{R}^{2}}+\frac{4}{5}\frac{\text{d}}{\text{R}}\Big)^{-\frac{3}{2}}\Bigg]$

For d << R, neglecting the factor$\frac{\text{d}^{2}}{\text{R}^{2}}$, we get:

≈$\frac{\mu_{0}\text{IR}^{2}}{2}\times\Big(\frac{5\text{R}^{2}}{4}\Big)^{-\frac{3}{2}}\bigg[\Big(1-\frac{4\text{d}}{5\text{R}}\Big)^{-\frac{3}{2}}+\Big(1+\frac{4\text{d}}{5\text{R}}\Big)^{-\frac{3}{2}}\Bigg]$

≈$\frac{\mu_{0}\text{IR}^{2}\text{N}}{2\text{R}^{3}}\Big(\frac{4}{5}\Big)^{\frac{3}{2}}\bigg[\Big(1-\frac{4\text{d}}{5\text{R}}\Big)^{\frac{3}{2}}+\Big(1+\frac{4\text{d}}{5\text{R}}\Big)^{\frac{3}{2}}\Bigg]$

$\text{B}=\Big(\frac{4}{5}\Big)^{\frac{3}{2}}\frac{\mu_{0}\text{IN}}{\text{R}}=0.72\Big(\frac{\mu_{0}\text{IN}}{\text{R}}\Big)$

Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.

1. If the wire intersects the axis, then the length of the wire is the diameter of the cyhndrtcal region.

Thus, I= 2r = 0.2 m

Angle between magnetic field and current, $\theta$ = 90°

Magnetic force acting on the wire is given by the relation,

$\text{F}=\text{BIl}\sin\theta$
$=1.5\times7\times0.2\times\sin90^{o}$

$=2.1\ \text{N}$

$\text{F}=\frac{\mu_{0}\text{I}^{2}}{2\pi\text{r}}$

Where,

$\mu_{0}=\text{Permeability of free space}=4\pi\times10^{-7}\text{T m A}^{-1}$

$\therefore\text{F}=\frac{4\pi\times10^{-7}\times(300)^{2}}{2\pi\times0.015}$

$=1.2\ \text{N/m}$

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

2. New length of the wire after turning it to the Northeast-Northwest direction can be given as:

$\text{l}_{1}=\frac{\text{l}}{\sin\theta}$

Angle between magnetic field and current, $\theta$ = 45°
Force on the wire,

$\text{F}=\text{BIl}_{1}\sin\theta$

$=\text{BIl}$

$=1.5\times7\times0.2$

$=2.1\ \text{N}$

Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle $\theta$  because/ $\sin\theta$ is fixed.

3. The wire is lowered from the axis by distance, d = 6.0 cm Let l₂ be the new length of the wire.

$\therefore\Big(\frac{\text{l}_{2}}{2}\Big)^{2}=4(\text{d+r})$

$=4(10+6)=4(16)$

$\therefore=8\times2=16\text{cm}=0.16\text{m}$

Magnetic force exerted on the wire,

$\text{F}_{2}=\text{BIl}_{2}$

$=1.5\times7\times0.16$

$=1.68\ \text{N}$

Hence, a force of 1.68 N acts in a vertically downward direction on the wire.

1. Torque, $\vec{\tau} =\text{I}\vec{\text{A}}\times\vec{B}$
   From the given figure, it can be observed that A is normal to the y-z plane and Bis directed along the z-axis.

$\therefore\tau=12\times(50\times10^{-4})\hat{\text{i}}\times0.3\hat{\text{k}}$

$=-1.8\times10^{-2}\hat{\text{j}}\ \text{Nm}$

The torque is 1.8 × I0^-2 N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.

2. This case is similar to case (a). Hence, the answer is the same as (a).
3. Torque $\vec{\tau} =\text{I}\vec{\text{A}}\times\vec{\text{B}}$

From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.

$\therefore\tau=-12\times(50\times10^{-4})\hat{\text{i}}\times0.3\hat{\text{k}}$

$=-1.8\times10^{-2}\hat{\text{j}}\ \text{Nm}$

The torque is 1.8 × 10^-2 N m along the negative x direction and the force is zero.

4. Magnitude of torque is given as:

$|\tau|=\text{lAB}$

$=12\times50\times10^{-4}\times0.3$

$=1.8\times10^{-2}\ \text{N m}$

Torque is 1.8 × 10^-2 N m at an angle of 240° with positive x direction. The force is zero.

5. Torque $\vec{\tau} =\text{I}\vec{\text{A}}\times\vec{B}$

$=(50\times10^{-4}\times12)\hat{\text{k}}\times0.3\hat{\text{k}}$

$=0$

Hence, the torque is zero. The force is also zero.

6. Torque $\vec{\tau} =\text{I}\vec{\text{A}}\times\vec{B}$

$=(50\times10^{-4}\times12)\hat{\text{k}}\times0.3\hat{\text{k}}$

$=0$

Hence, the torque is zero. The force is also zero.

In case (e), the direction of $I\vec{\text{A}}$ and $\vec{\text{B}}$ is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable. whereas, in case (f), the direction of $I\vec{\text{A}}$ and $\vec{\text{B}}$ is opposite. The angle between them is 180°. If disturbe, it does not come back to its original position. Hence, its equilibriun is unstable.

Charge on the electron, e = 1.6 × 10^-19 C

Mass of the electron, m = 9.1 × 10^-31 kg

Potential difference, V = 2.0 kV = 2 × 10³ V

Thus, kinetic energy of the electron = eV

$\Rightarrow\text{eV}=\frac{1}{2}\text{mv}^{2}$

$\text{v}=\sqrt\frac{2\text{ev}}{\text{m}}.........(1)$

Where,

v = velocity of the electron

1. Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

Magnetic force on the electron is given by the relation,

$\text{B ev}$

$\text{Centripetal force}=\frac{\text{mv}^{2}}{\text{r}}$

$\therefore\text{Bev}=\frac{\text{mv}^{2}}{\text{r}}$

$\text{r}=\frac{\text{mv}}{\text{Be}}.........(2)$

From equations (1) and (2), we get

$\text{r}=\frac{\text{m}}{\text{Be}}\Big[\frac{2\text{ev}}{\text{m}}\Big]^{\frac{1}{2}}$

$=\frac{9.1\times10^{-31}}{0.15\times1.6\times10^{-19}}\times\Big(\frac{2\times1.6\times10^{-19}\times2\times10^{3}}{9.1\times10^{-31}}\Big)^{\frac{1}{2}}$

$=100.55\times10^{-5}$

$=1.01\times10^{-3}\text{m}$

$=1\text{mm}$

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

2. When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,

$\text{v}_{1}=\text{v}\sin\theta$

From equation (2), we can write the expression for new radius as: $\text{r}_{1}=\frac{\text{mv}_{1}}{\text{Be}}$

$=\frac{\text{mv}\sin\theta}{\text{Be}}$

$=\frac{9.1\times10^{-31}}{0.15\times1.6\times10^{-19}}\times\Big(\frac{2\times1.6\times10^{-19}\times2\times10^{3}}{9\times10^{-31}}\Big)^{\frac{1}{2}}\times\sin30^{\circ}$

$=0.5\times10^{-31}\text{m}$

$=0.5\text{mm}$

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction. 

1. If magnetic compass of dipole moment $\overrightarrow{\text{m}}$ is placed at angle $\theta$ in uniform magnetic field, and released it experiences a restoring torque.

$\overrightarrow{\tau} = - \text{magnetic force}\times\text{ perpendicular distance}$

$ = - |\text{mB}|.( 2 \text{ a }\sin\theta)$

$\overrightarrow{\tau} = - \overrightarrow{\text{m}}\times\overrightarrow{\text{B}}$

where m = pole strength

$|\tau| = - \text{m}|\text{B}|.\sin\theta $

In equilibrium, the equation of motion,

$\Rightarrow\text{I}\frac{\text{d}^{2}\theta}{\text{dt}^{2}} = -|\text{m}||\text{B}|\theta$

$\Rightarrow\frac{\text{d}^{2}\theta}{\text{dt}^{2}} = - \frac{|\text{M}||\text{B}|}{\text{I}}\theta$ $\text{(For small angle} \sin \theta \approx \theta)$

$\Rightarrow\frac{\text{d}^{2}\theta}{\text{dt}^{2}} = - \frac{\text{|M}||\text{B}|}{\text{I}}\theta$

$\Rightarrow\frac{\text{d}^{2}\theta}{\text{dt}^{2}} = - \bigg(\frac{\text{MB}}{\text{I}}\bigg)\theta$

Since $\frac{\text{d}^{2}\theta}{\text{dt}^{2}}\propto\theta$

It represents the simple harmonic motion with angular frequency

$\omega^{2} = \frac{\text{|M|}\text{B}|}{\text{I}}$

$\Rightarrow\text{T} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{\text{I}}{\text{MB}}}$

2. If compass needle orients itself with its axis vertical at a place, then
   1. B_H = 0 because B_V = |B|
   2. Angle of dip $\delta = 90^{o}$

$\tan\delta = \frac{\text{B}_{V}}{\text{B}_{H}} = \infty$

$\Rightarrow\text{Angle }\delta = 90^{o}.$

Concept - It is possible only on magnetic north or south poles.

1. (d) Is a deflection instrument which gives a deflection when a current flows through its coil.

Explanation:

A moving coil galvanometer is a sensitive instrument which is used to measure a deflection when a current flows th rough its coil.

2. (d) Poles are cylindrically cut.

Explanation:

Uniform field is made radial by cutting pole pieces cylindrically.

3. (b) Directly proportional to the number of turns in the coil.

Explanation:

The deflection in a moving coil galvanometer, $\phi=\frac{\text{NAB}}{\text{K}}.\text{I}$ or $\phi\propto\text{N,}$ N, where N is number of turns in a oil, Bis magnetic field and A is area of cross-section.

4. (d) NABI

Explanation:

The deflecting torque acting on the coil. $\tau_{\text{deflection}}=\text{NIAB}$

5.  (b) Torsional constant of spring.

Explanation:

Current sensitivity of galvanometer,

$\frac{\phi}{\text{I}}=\text{S}_\text{i}=\frac{\text{NBA}}{\text{K}}$

Hence, to increase (current sensitivity) S;, (torsional constant of spring) k must be decrease.

1. (d) $9975\Omega$

Explanation:

A galvanometer can be converted into a voltmeter of given range by connecting a suitable high resistance R in series of galvanometer, which is given by,

$\text{R}=\frac{\text{V}}{\text{I}_\text{g}}-\text{G}=\frac{100}{10\times10^{-3}}-25=10000-25=9975\Omega$

2. (c) $15000\Omega$

Explanation:

An ideal voltmeter should have a very high resistance.

3. (c) $990\Omega$

Explanation:

Resistance of voltmeter $=\frac{25}{25\times10^{-3}}=1000\Omega$

$\therefore\text{X}=1000-10=990\Omega$

4. (d) A high resistance R is connected in series with MCG.

Explanation:

To convert a moving coil galvanometer into a voltmeter, it is connected with a high resistance in series. The voltmeter is connected in parallel to measure the potential difference. As the resistance is high, the voltmeter itself does not consume current.

5. (d) Infinity.

Explanation:

The resistance of an ideal voltmeter is infinity.

1. (c) By accelerating them through electric and magnetic field.

Explanation:

In mass spectrometer, the ions are sorted out by accelerating them through electric and magnetic field.

2.  (c) $\frac{\text{mv}}{\text{qB}_0}$

Explanation:

As $\frac{\text{mv}^2}{\text{r}}=\text{qvB}_0\ \therefore\text{r}=\frac{\text{mv}}{\text{qB}_0}$

3.  (b) ct-particle

Explanation:

As radius $\text{r}\propto\frac{\text{m}}{\text{q}}.$

$\therefore$ r will be maximum for a - particle.

4.  (b) $\frac{\text{qbr}}{\text{E}}$

Explanation:

Here, $\text{r}\propto\frac{\text{mv}}{\text{qB}_0}.$ or $\text{m}=\frac{\text{rqB}_0}{\text{v}}$

As $\text{v}=\frac{\text{E}}{\text{B}},\ \therefore\text{m}=\frac{\text{qB}_0\text{Br}}{\text{v}}$

5.  (c) Electric and magnetic force balance each other.

Explanation:

From the relation $\text{v}=\frac{\text{E}}{\text{B}},$ it is clear electric and magnetic force balance each other.

1. (b) 2r₀

Explanation:

As, $\text{r}_0=\frac{\text{mv}}{\text{qB}}\Rightarrow\text{r}'=\frac{\text{m}(2\text{v}_0)}{\text{qB}}=2\text{r}_0$

2.  (c) T₀

Explanation:

As, $\text{T}=\frac{2\pi\text{m}}{\text{qB}}$

Thus, it remains same as it is independent of velocity.

3. (b) -4ms^-2

Explanation:

As, $\text{F}\bot\text{B}$

Hence, $\text{a}\bot\text{B}$

$\therefore\vec{\text{a}}.\vec{\text{B}}=0$

$\Rightarrow(\text{x}\hat{\text{i}}+2\hat{\text{j}}=0)$

$2\text{x}+8=0\Rightarrow\text{x}=-4\text{ms}^{-2}$

4. (c) Helical.

If the charged particle has a velocity not perpendicular to $\vec{\text{B}},$ then component of velocity along $\vec{\text{B}}$ remains unchanged as the motion along the $\vec{\text{B}}$ will not be affected by $\vec{\text{B}}.$

Then, the motion of the particle in a plane perpendicular to $\vec{\text{B}},$ is as before circular one. Thereby, producing helical motion.

5.  (d) Zero

 Explanation:

The force on electron $\text{F}=\text{qv}\text{B}\sin\theta$

As the electron is moving parallel to B

So, $\theta=0^\circ\Rightarrow\text{qv}\text{B}\sin\theta$

1. (b) 8 × 10^-4T

Explanation:

As, $\text{B}=\frac{\mu_0\text{nl}}{2}=\frac{(4\pi\times10^{-7})\times800\times1.6}{2}$

$=8\times10^{-4}\text{T}$

2.  (c) The magnetic field at the centre, inside an ideal solenoid is almost twice that at the ends.

Explanation:

Magnetic field at one end of a solenoid carrymg current is $\text{B}=\frac{\mu_0\text{nl}}{2}$ Magnetic field inside the solenoid is uniform and is given by $\text{B}_\text{c}=\mu_0\text{nl}$

3.  (d) $\mu_0\text{ nI}(\text{1}+\chi)$

Explanation:

Magnetic field inside a long solenoid with an iron core inside it is $\text{B}=\mu\text{nl}$

But $\mu=\mu_0(1+\chi)\therefore\text{B}\mu_0(1+\chi)\text{nl}$

4.  (c) A​​​​​​along the axis of solenoid.

Explanation:

A solenoid of length l and having n turns carries a current I in anticlockwise direction. The magnetic field is $\frac{\mu_0\text{nl}}{\text{l}}.$ Its direction will be along the axis of solenoid.

5.  (d) Introducing a medium of higher permeability.

1.  (a) Remains stationary.

Explanation:

For stationary electron, $\vec{\text{v}}=0$

$\therefore$ Force on the electron is, $\vec{\text{F}}_\text{m}=-\text{e}(\vec{\text{v}}\times\vec{\text{B}})=0$

2. (d) The proton will continue to move with velocity v along the axis.

Explanation:

Force on the proton, $\vec{\text{F}}_\text{B}=\text{e}(\vec{\text{v}}\times\vec{\text{B}})$

Since, $\vec{\text{v}}$ is parallel to $\vec{\text{B}}$

$\therefore\vec{\text{F}}_\text{B}=0$

Hence proton will continue to move with velocity v along the axis of solenoid.

3. (b) The particle is moving and magnetic field is perpendicular to the velocity.

Explanation:

Magnetic force on the charged particle q is,

$\vec{\text{F}}_\text{m}=\text{q}(\vec{\text{v}}\times\vec{\text{B}})$ or $\text{F}_\text{m}=\text{qv}\text{B}\sin\theta$

where $\theta$ is the angle between $\vec{\text{v}}$ and $\vec{\text{B}}.$

Out of the given cases, only in case (b) it will experience the force while in other cases it will experience no force.

4. (a) In z-y plane.

Explanation:

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

$=\text{q[}2\vec{\text{i}}\times(\vec{\text{i}}+2\vec{\text{j}}+3\vec{\text{k}})]=(4\text{q})\vec{\text{k}}-(6\text{q})\vec{\text{j}}$

5.  (c) Both electric and magnetic field.

Explanation:

When an electric charge is moving both electric and magnetic fields are produced, whereas a static charge produces only electric field. 

1. When switch S is open

2T cos 30° = mg

$\Rightarrow\text{T}=\frac{\text{mg}}{2\cos30^\circ}$

$=\frac{200\times10^{-3}\times9.8}{2\sqrt{\frac{3}{2}}}=1.13$

2. When the switch is closed and a current passes through the circuit = 2A

Then,

$\Rightarrow2\text{T}\cos30^\circ=\text{mg}+\text{ilB}$

$=200\times10^{-3}9.8+2\times0.2\times0.5$

$=1.96+0.2=2.16$

$\Rightarrow2\text{T}=\frac{2.16\times2}{\sqrt{3}}=2.49$

$\Rightarrow\text{T}=\frac{2.49}{2}=1.245\approx1.25$

1. (d) Any one of (i), (ii) and (iii)

2.  (a) B will have smaller radius of curvature than that of A.

Explanation:

Using, $\text{qv}\text{B}\sin\theta=\frac{\text{mv}^2}{\text{r}}$

$\text{r}\propto\frac{1}{\sin\theta}$ for the same values of m, v, q and B

$\therefore\frac{\text{r}_\text{A}}{\text{r}_\text{B}}=\frac{\sin90^\circ}{\sin30^\circ}=2$ or $\text{r}_\text{A}=2\text{r}_\text{B}$ or $\text{r}_\text{B}<\text{r}_\text{A}$

3. (d) 0.5mm

Explanation:

The radius of the helical path of the electron in the uniform magnetic field is:

$\text{r}=\frac{\text{mv}\perp}{\text{eB}}=\frac{\text{mv}\sin\theta}{\text{eB}}=\frac{(2.4\times10^{-23}\text{kg m/ s})\times\sin30^\circ}{(1.6\times10^{-19}\text{C})\times0.15\text{T}}$

$=5\times10^{-4}\text{m}=0.5\times10^{-3}\text{m}=0.5\text{mm}$

4. (c) 0.157m

Explanation:

Here, $\vec{\text{B}}=8.35\times10^{-2}\hat{\text{i}}\text{T}$

$\vec{\text{v}}=2\times10^5\hat{\text{i}}+4\times10^5\hat{\text{j}}\frac{\text{m}}{\ \text{s}},$

$\text{m}=1.67\times10^{-27}\text{kg}$

Pitch of the helix (i.e., the linear distance moved along the magnetic field in one rotation) is given by,

Pitch of the helix $=\frac{2\pi\text{mv}_{||}}{\text{qB}}$

$=\frac{2\times3.14\times1.67\times10^{27}\times2\times10^5}{1.6\times10^{-19}\times8.35\times10^{-2}}=0.157\text{m}$

5.  (b) $\frac{\text{qB}}{2\pi\text{m}}$

 Explanation:

Period ofrevolution

$\text{T}=\frac{2\pi\text{R}}{\text{v}\sin\theta}\Rightarrow\text{T}\frac{2\pi(\frac{\text{mv}\sin\theta}{\text{qB}})}{\text{v}\sin\theta}\Rightarrow\text{T}=\frac{2\pi\text{m}}{\text{qB}}$

$\therefore$ Frequency, $\upsilon=\frac{1}{\text{T}}=\frac{\text{qB}}{2\pi\text{m}}$

1. (a) Zero

Explanation:

Torque on a current carrying loop in magnetic field, $\tau=\text{IBA}\sin\theta$

Here, $\text{I}=10\text{A},\text{B}=0.1\text{T},\text{A}=1\text{cm}^2=10^{-4}\text{m}^2,\theta=0^0$

$\therefore\tau=10\times0.1\times10^{-4}\sin0^0=0$

2.  (a) $\text{m}\propto\omega$

Explanation:

Magnetic moment, $\text{M}=\text{IA}=\text{I}(\pi\text{r}^2)=\frac{\text{q}}{T}\times\pi\text{r}^2$

As, $\omega=\frac{2\pi}{\text{T}}\ \therefore\text{ M}=\frac{\text{q}\omega\text{r}^2}{2}\ \text{ or }\text{ M}\propto\omega$

3. (b) Can be in equilibrium in two orientations, one stable while the other is unstable.

Explanation:

When a current loop is placed in a magnetic field it experiences a torque. It is given by,

$\vec{\tau}=\vec{\text{M}}\times\vec{\text{B}}$

Where M is the magnetic moment of the loop and B is the magnetic field.

or $\tau=\text{MB}\sin\theta$ where $\theta$ is angle between M and B When $\vec{\text{M}}$ and $\vec{\text{B}}$ are parallel (i.e. $\theta$ = 0⁰⁾ the equilibrium is stable and when they are antiparallel (i.e. $\theta$) the equilibrium is unstable.

4. (d) $\text{r}^2$

 Explanation:

Magnetic moment, $\text{M}=\text{NIA}=\text{NI}\pi\text{r}^2\text{ i.e., }\text{M}\propto\text{r}^2$

5. (a) Force is non-zero.

1. (a) $\frac{\mu_0\text{I}}{2\pi\text{r}}$

Explanation:

Magnetic field due to a long current carrying wire at r.

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

2.  (d) $\frac{\mu_0\text{Ir}}{2\pi\text{R}^2}$

Explanation:

Let I' be the current in region r < R

Then, $\text{I}'=\frac{\text{I}}{\pi\text{R}^2}\pi(\text{r}^2)$ or $\text{I'}=\frac{\text{Ir}^2}{\text{R}^2}$

So, magnetic field, $\text{B}=\frac{\mu_0\text{I}'}{2\pi\text{r}}=\frac{\mu_0\text{Ir}^2}{2\pi\text{R}^2\text{r}}=\frac{\text{Ir}}{\text{R}^2}$

3.  (a) Proton

Explanation:

Magnetic field due to a long straight wire of radius a carrying current I at a point distant rfrom the centre of the wire is given as follows

$\text{B}=\frac{\mu_0\text{Ir}}{2\pi\text{a}^2}\ \ \ \ \text{for}\text{ r}<\text{a}$

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

$\text{B}=\frac{\mu_0\text{I}}{2\pi\text{r}}\ \ \ \ \text{for}\text{ r}>\text{a}$

The variation of magnetic field B with distance r from the centre of wire is shown in the figure.

4.  (d) $1$

Explanation:

Let the magnetic fields due to a long straight wire of radius Rcarrying a steady current lat a distance r from the centre of the wire are.

$\text{B}_1\frac{\mu_0\text{Ir}}{2\pi\text{R}^2}\ \ \ (\text{for }\text{r}<\text{R})$

and $\text{B}_2\frac{\mu_0\text{I}}{2\pi\text{R}}\ \ \ (\text{for r}>\text{R})$

So, the magnetic field at $\text{r}=\frac{\text{R}}{2}$ is $\text{B}_1=\frac{\mu_0\text{I}}{2\pi\text{r}^2}(\frac{\text{R}}{2})=\frac{\mu_0\text{I}}{4\pi\text{R}}$

and at r = 2 is $\text{B}_2=\frac{\mu_0\text{I}}{2\pi(2\text{R})}=\frac{\mu_0\text{I}}{4\pi\text{R}}$

$\therefore$ Their corresponding ratio is $\frac{\text{B}_1}{\text{B}_2}=\frac{\frac{\mu_0\text{I}}{4\pi\text{R}}}{\frac{\mu_0\text{I}}{4\pi\text{R}}}=1$

5.  (c) Zero at any point inside the pipe.

1. (c) Perpendicular to both $\text{d}\vec{\text{l}}$ and $\vec{\text{r}}$

Explanation:

According to Biot-Savart's law, the magnetic induction due to a current element is given by,

$\text{d}\vec{\text{B}}=\frac{\mu_0}{4\pi}\frac{\text{Id}\vec{\text{l}\times\text{r}}}{\text{r}^3}$

This is perpendicular to both $\text{d}\vec{\text{l}}$ and $\vec{\text{r}}.$

2. (b) Decreases as

Explanation:

From Biot-savart's law, $\text{dB}=\frac{\mu_0}{4\pi}\frac{\text{Idl}}{\text{r}^2}\text{i.e. dB}\propto\frac{1}{\text{r}^2}$

3. (c) $\text{Zero}$

Explanation:

$\text{B}=\frac{\mu_0}{2\pi}.\frac{\text{i}}{\text{r}}-\frac{\mu_0}{2\pi}.\frac{\text{i}}{\text{r}}=0$

4. (a) The magnetic fields are equal.

5. (b) Ampere's circuital law.

Explanation:

Biot-Savart law can be expressed alternatively as Ampere circuital law.