Electromagnetism PYQ — Page 60

An electric field $\vec{E}=(25 \hat{i}+30 \hat{j})N{C}^{-1}$ exists in a region of space. If the potential at the origin is taken to be zero then the potential at $x=2m$, $y=2m$ is:

A $25\mathrm{cm}$ long solenoid has the radius $2\mathrm{cm}$ and $500$ turns. It carries a current of $15A$. If it is equivalent to a magnet of the same size and magnetization $\vec{M} (\frac{Magnetic moment}{volume}),$then $|\vec{M}|$ is:

 Two long currents carrying thin wires, both with current $I$, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ' $\theta$ ' with the vertical. If wires have a mass $\lambda$ per unit length then the value of $I$ is: ($g=$ gravitational acceleration)

A wire of length $L=20 cm$ is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges $+Q$ and $-Q$ as shown in the figure. The magnitude of the charge on each half is $|Q|={10}^{3}{\epsilon }_{0}$, where ${\epsilon }_{0}$ is the permittivity of free the space. The net electric field at the centre $O$ is 

A $10 \text{V}$ battery with internal resistance $\text{1} \Omega$ and a $15 \text{V}$ battery with internal resistance $0.6 \Omega$ are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to: 

 In the circuit shown, the current in the $1\Omega$ resistor is:

An inductor $( L=0.03 \text{H} )$ and a resistor $(R=0.15 \text{kΩ})$ are connected in series to a battery of $15 \text{V}$ E.M.F. in a circuit shown below. The key ${K}_{1}$ has been kept closed for a long time. Then at $t=0$, ${K}_{1}$ is opened and key ${K}_{2}$ is closed simultaneously. At $t=1 \text{ms}$ , the current in the circuit will be : $(\text{Take},{e}^{5}\approx 150)$ 

For plane electromagnetic waves propagating in the $+z$-direction, which one of the following combinations gives the correct possible direction for $\vec{E}$ and $\vec{B}$ field respectively?

A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of $30\mathrm{cm}$ from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in ${\mathrm{Am}}^{2}$ is close to: (Given $\frac{{\mu }_{0}}{4\pi }={10}^{-7}$ in SI units and ${B}_{H}=$ Horizontal component of earth's magnetic field $=3.6\times {10}^{-5}$ Tesla.)

A proton (mass $m$) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space by width $d$. If $\alpha$ be the angle of deviation of proton from the initial direction of motion (see figure), the value of $\mathrm{sin}⁡\alpha$ will be: 

When the current in a coil changes from $\text{5 A}$ to $\text{2 A}$ in $\text{0}\text{.1 s}$, an average voltage of $\text{50} \text{V}$ is produced. The self-inductance of the coil is

A thin disc of radius $b=2a$ has a concentric hole of radius $a$ in it (see figure). It carries uniform surface charge $\sigma$ on it. If the electric field on its axis at a height $\text{h}(\text{h}<<\text{a})$ from its centre is given as $\text{C}\text{h}$ then the value of $C$ is 

Shown in the figure are two point charges $+Q$ and $-Q$ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If ${\sigma }_{1}$ is the surface charge on the inner surface and ${Q}_{1}$ net charge on it and ${\sigma }_{2}$ the surface charge on the outer surface and ${Q}_{2}$ net charge on it then: 

The mid points of two small magnetic dipoles of length $d$ in end-on positions, are separated by a distance $x$$(x\gg d)$. The magnitude of force between them is proportional to ${x}^{-n}$ where $n$ is : 

Assume that an electric field $\vec{\text{E}} = 3 0 { x⁡ }^{2} \hat{ i⁡ }$ exists in space. Then the potential difference ${\text{V}}_{\text{A}} - {\text{V}}_{\text{O}}$ , where V$_{O}$ is the potential at the origin and V$_{A}$ the potential at x = 2 m is :

An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $\mathrm{z}$-axis. The amplitude of electric field is $4 \mathrm{~V} / \mathrm{m}$. If $\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-$ $\mathrm{m}^2$, then average energy density of electric field will be:

A positive charge ' $\mathrm{q}$ ' of mass ' $\mathrm{m}$ ' is moving along the $+x$ axis. We wish to apply a uniform magnetic field B for time $\Delta t$ so that the charge reverses its direction crossing the $\mathrm{y}$ axis at a distance $\mathrm{d}$. Then:

In the circuit shown, current (in A) through $50 \mathrm{~V}$ and $30 \mathrm{~V}$ batteries are, respectively: 

A parallel plate capacitor is made of two plates of length 1 , width $w$ and separated by distance $d$. A dielectric slab (dielectric constant $\mathrm{K}$ ) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $\mathrm{F}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}$ where $\mathrm{U}$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is: 

When the rms voltages ${V}_{L} , {V}_{C}$ and ${V}_{R}$ are measured respectively across the inductor $L,$ the capacitor $C$ and the resistor $R$ in a series $LCR$ circuit connected to an $AC$ source, it is found that the ratio ${V}_{L} :{V}_{C} :{V}_{R} =1:2:3.$ If the rms voltage of the $AC$ source is $100 V,$ then ${V}_{R}$ is close to :

Match List I (Wavelength range of electromagnetic spectrum) with List II (Method of production of these waves) and select the correct option from the options given below the lists.<table class="pyq-table"><tbody><tr><th></th><th>List I</th><th></th><th>List II</th></tr><tr><td>(a)</td><td>$700\mathrm{nm}$ to $1\mathrm{mm}$</td><td>(i)</td><td>Vibration of atoms and molecules.</td></tr><tr><td>(b)</td><td>$1\mathrm{nm}$ to $400\mathrm{nm}$</td><td>(ii)</td><td>Inner shell electrons in atoms moving from one energy level to a lower level.</td></tr><tr><td>(c)</td><td>$<{10}^{-3}\mathrm{nm}$</td><td>(iii)</td><td>Radioactive decay of the nucleus.</td></tr><tr><td>(d)</td><td>$1\mathrm{mm}$ to $0.1m$</td><td>(iv)</td><td>Magnetron valve.</td></tr></tbody></table>

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150N/C$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given : ${\in }_{\text{O}} = \text{8.85} \times 1 {0}^{ - 1 2 } {\text{ C}}^{2} / {\text{N-m}}^{2} \text{, } {\text{R}}_{\text{E}} = \text{6.37} \times 1 {0}^{6} \text{m}$]

The electric field in a region of space is given by, $\vec{E}={E}_{0}\hat{i}+2{E}_{0}\hat{j}$ where ${E}_{0}=100N{C}^{-1}$. The flux of this field through a circular surface of radius $0.02m$ parallel to the $Y‐Z$ plane is nearly

The coercivity of a small magnet, where the ferromagnet gets demagnetised is $3\times {10}^{3}A/m$. The current required to be passed in a solenoid of length $10\mathrm{cm}$ and number of turns $100$, so that the magnet gets demagnetised when inside the solenoid is