Physics 4 Marks Questions

1. (b) $\frac{\pi}{3}$

Explanation:

$\tan\theta=\frac{\text{B}_\text{V}}{\text{B}_\text{H}}$ and $\text{B}_\text{H}=\frac{\text{B}_\text{V}}{\sqrt{3}}$

$\therefore\tan\theta=\sqrt{3}\text{ i.e. }\theta=\frac{\pi}{3}$

2. (b) Magnetic declination.

Explanation:

The angle between the true geographic north and the north shown by a compass needle is called as magnetic declination or simply declination.

3. (d) 90º, 0º 

Explanation:

Since angle of dip at a place is defined as the angle $\delta,$ which is the direction of total intensity of earth's magnetic field B makes with a horizontal tine in magnetic meridian,

At poles $\text{B}=\text{B}_\text{V}$ and $\text{B}_\text{V}=\text{B}\sin\delta\therefore\sin\delta=1\Rightarrow\delta=90^\circ$

At equator $\text{B}=\text{B}_\text{H}$ and $\text{B}_\text{H}=\text{B}\cos\delta$

$\therefore\cos\delta=1\Rightarrow\delta=-0^\circ.$

4. (a) Will become rigid showing no movement.

Explanation:

A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It will stay in any position as the horizontal component of earth's magnetic field becomes zero at the geomagnetic pole.

5. (c) Magnetic dip increases as we move away from the equator towards the magnetic pole.

Explanation: 

At equator, $\delta=0^\circ$

At poles, $\delta=90^\circ$

$\therefore\delta$ increases as we move from equator towards poles.

$\text{d}=10\text{cm}=0.1\text{m}$

We know

$\frac{\text{M}}{\text{B}}=\frac{4\pi}{\mu_0}\frac{(\text{d-}^2\ell^2)^2}{2\text{d}}\tan\theta$

$=\frac{4\pi}{\mu_0}\times\frac{\text{d}^4}{2\text{d}}\tan\theta$ [As the magnet is short]

$=\frac{4\pi}{4\pi\times10^{-7}}\times\frac{(0.1)^3}{2}\times\tan37^\circ$

$=0.5\times0.75\times1\times10^{-3}\times10^7$

$=3.75\times10^3\text{A-m}^2\text{T}^{-1}$

1. If the point at end-on postion.

$\text{B} =\frac{\mu_02\text{M}}{4\pi\text{ d}^3}\Rightarrow2\times10^{-4}=\frac{10^{-7}\times2\text{M}}{(10^ {-1})^3}$

$\Rightarrow\frac{2\times10^{-4}\times10^{-3}}{10^{-7}\times2}=\text{M}\Rightarrow\text{M}=1\text{Am}^2$

2. If the point is at broad-on position.

$\frac{\mu_0\text{ M }}{4\pi\text{ d}^3}\Rightarrow2\times10^{-4}=\frac{10^{-7}\times\text{M}}{(10^{-1})^3}\Rightarrow\text{m}=2\text{Am}^2$

1. (b) Net magnetic moment per unit volume.

2. (d) Magnetic permeability - Henry m

Explanation:

Magnetic penneability- Henry m^-1.

3. (d) 5 × 10⁵A/ m

Explanation:

Given, l = 3cm, A= 2cm ², M = 3A m² intensity of magneusanon $=\frac{\text{M}}{\text{lA}}=\frac{3}{3\times10^{-2}\times2\times10^{-4}}$

4. (b) 2.5 × 10⁵Am^-1

Explanation:

Here, n = 500 turns/ m

$\text{I}=\text{lA},\ \mu^{\text{r}}=500$

Magnetic intensity, $\text{H}=\text{nl}=500\text{m}^{-1}\times1\text{A}=500\text{Am}^{-1}$

As $\mu^{\text{r}}=1+\chi$ or $\chi=(\mu_\text{r}-1)$

Magnetisation, $\text{M}=\chi\text{H}$

$=(\mu_\text{r}-1)\text{H}=(500-1)\times500\text{Am}^{-1}$

$=\text{2.495}\times106{5}\text{Am}^{-1}\approx2.5\times10^{5}\text{Am}^{-1}$

5. (a) 5999

Explanation: 

Relative permeability of iron, $\mu_{\text{r}}=6000$

Magnetic susceptibility $\chi_{\text{m}}=\mu_{\text{r}}-1=5999.$

1. (b) Case (ii) contradicts Gauss's law for magnetic fields.

Explanation:

According to Gauss's law in magnetism $\oint\vec{\text{B}}.\text{d}\vec{\text{s}}=0,$ which implies that number of magnetic field tines entering the Gaussian surface is equal to the number of magnetic field lines leaving it. Therefore, case (ii) is not possible.

2. (a) $\text{Zero}$

Explanation:

The net magnetic flux through a closed surface will be zero, $\oint\vec{\text{B}}.\text{d}\vec{\text{s}}=0,$ because there are no magnetic monopoles.

3. (b) $\text{Zero}$

4. (c) There are abundant sources or sinks of the magnetic field inside a closed surface.

Explanation:

Gauss's law indicates that there are no sources or sinks of the magnetic field inside a closed surface. ln other words, there are no free magnetic charges.

5. (d) Equal to its magnetic flux through that surface.

Explanation: 

The surface integral of a magnetic field over a surface gives magnetic flux through that surface.

$\theta=\tan^{-1}\sqrt{2}\Rightarrow\tan\theta=\sqrt{2}\Rightarrow2=\tan^2\theta$

$\Rightarrow\tan\theta=2\cot\theta\Rightarrow\frac{\tan\theta}{2}=\cot\theta$

We know $\frac{\tan\theta}{2}=\tan\propto$

Comparing we get, $\tan\propto=\cot\theta$

 $\tan\propto=(90-\theta)$

$\propto=90-\theta$

 $\theta +\propto=90$

Hence magnetic field due to the dipole is $\perp\text{r}$ to the magnetic axis.

1. (c) 10^-5T

2. (b) East- West

3. (c) The value of V is zero.

Explanation:

At equator vertical component of magnetic fields is zero.

4. (d) 45º

Explanation:

Given, $\text{V}=\text{H}$

$\therefore\tan\delta=\frac{\text{V}}{\text{H}}=1$ or $\delta=45^\circ$

5.  (b) $\frac{2\text{H}}{\sqrt{3}}$

Explanation: 

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

1. (a) a

Explanation:

It can be seen that slop of curve for wire a is greater th an wire c. 

2. (b) Less than in wire c.

Explanation:

Inside the wire

$\text{B(r)}=\frac{\mu_0}{2\pi}\frac{\text{I}}{\text{R}^2}\text{r}\Rightarrow\frac{\text{dB}}{\text{dr}}=\frac{\mu_0}{\text{dr}}=\frac{\text{I}}{\text{R}^2}\text{r}$

$\text{i.e. slope}\propto\frac{\text{I}}{\pi\text{R}^2}\propto\text{Current density}$

3. (c) c

Explanation:

Wire c has the greatest radius.

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

5.  (a) Outside the cable.

$\text{dx}=10\sin30^\circ\text{cm}=5\text{cm}$

$\frac{\text{dV}}{\text{dx}}=\text{B}=\frac{0.1\times10^{-4}\text{T-m}}{5\times10^{-2}\text{m}}$

Since B is perpendicular to equipotential surface.

Here it is at angle 120° with (+ve) x-axis and B = 2 × 10^-4 T

$=-0.2\times10^{-3}\times0.5=-0.1\times10^{-3}\text{T-m}$

$\text{B}_\text{H}=\text{B}\cos60^\circ$

$\Rightarrow\text{B}=52\times10^{-6}=52\mu\text{T}$

$\text{B}_\text{v}=\text{B}\sin\delta=52\times10^{-6}\frac{\sqrt{3}}{2}$

$=44.98\mu\text{T} \approx45\mu\text{T}$