Electromagnetism PYQ — Page 54
JEE Main Physics — Electromagnetism previous year questions with solutions.
All Electromagnetism Questions (1703)
A parallel plate capacitor having capacitance $12 \mathrm{pF}$ is charged by a battery to a potential difference of $10 V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is
A moving coil galvanometer has a coil with $175$ turns and area $1 c{m}^{2}.$ It uses a torsion band of torsion constant ${10}^{-6} Nm/rad$ The coil is placed in a magnetic field B parallel to its plane. The coil deflects by ${1}^{o}$ for a current of $1 mA$ The value of B (in tesla) is approximately:
The self induced emf of a coil is $25$ volts. When the current in it is changed at uniform rate from $10A$ to $25A$ in $1s$ , the change in the energy of the inductance is:
As shown in the figure, two infinitely long, identical wires are bent by ${90}^{o}$ and placed in such a way that the segments $LP$ and $QM$ are along the $x$ - axis, while segments $PS$ and $QN$ are parallel to the $y$ - axis. If $OP=OQ=4cm,$ and the magnitude of the magnetic field at $O$ is ${10}^{-4} T,$ and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at $O$ will be $({\mu }_{0}=4\pi \times {10}^{-7}N{A}^{-2}):$ 
A moving coil galvanometer, having a resistance $G,$ produces full scale deflection when a current ${I}_{G}$ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to ${I}_{o}({I}_{0}>{I}_{g})$ by connecting a shunt resistance ${R}_{A}$ to it and (ii) into a voltmeter of range 0 to $V(V=G{I}_{0})$ by connecting a series resistance ${R}_{V}$ to it. Then,
The figure shows a square loop $L$ of side $5 cm$ which is connected to a network of resistances. The whole setup is moving towards the right with a constant speed of $1cm{s}^{-1}$ . At some instant, a part of $L$ is in a uniform magnetic field of $1T$ perpendicular to the plane of the loop. If the resistance of $L$ is $1.7 \Omega ,$ the current in the loop at that instant will be close to: 
Four equal point charges $Q$ each are placed in the $xy$ plane at $(0, 2),(4, 2),(4, -2)$ and $(0, -2)$ . The work required to put a fifth charge $Q$ at the origin of the coordinate system will be:
To verify Ohm’s law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained:  If ${V}_{o}$ is almost zero, identify the correct statement:
A proton and an $\alpha$- particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$ ) are accelerated from rest through a potential difference $V.$ If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii ${r}_{p}:{r}_{\alpha }$ of the circular paths described by them will be:
A galvanometer of resistance $100 \Omega$ has $50$ divisions on its scale and has sensitivity of $20\mu A/ division$ . It is to be converted to voltmeter with three ranges, of $0-2 V,0-10 V$ and $0-20 V.$ The appropriate circuit to do so is:
The bob of a simple pendulum has mass $2 g$ and a charge of $5.0 \mu C.$ It is at rest in a uniform horizontal electric field of intensity $2000 V/m$ At equilibrium, the angle that the pendulum makes with the vertical is: $(takeg=10 m/{s}^{2})$
Charges $-q$ and $+q$, located at $A$ and $B$, respectively, constitute an electric dipole. Distance $AB=2a$, $O$ is the mid point of the dipole and $OP$ is perpendicular to $AB$. A charge $Q$ is placed at $P$ where $OP=y$ and $y\gg 2a$. The charge $Q$ experiences an electrostatic force $F$. If $Q$ is now moved along the equatorial line to $P'$ such that $OP'=(\frac{y}{3})$ the force on $Q$ will be close to$(\frac{y}{3}\ll 2a)$ 
A current of $5 A$ passes through a copper conductor (resistivity $=1.7\times {10}^{-8} \Omega m$ ) of radius of cross-section $5 mm$ . Find the mobility of the charges if their drift velocity is $1.1\times {10}^{-3} {\mathrm{ms}}^{-1}$ .
In the circuit shown, find $C$ if the effective capacitance of the whole circuit is to be $0.5 \mu F.$ All values in the circuit are in $\mu F.$ 
Shown in the figure is a shell made of a conductor. It has inner radius $a$ and outer radius $b$, and carries charge $Q$ . At its centre a dipole $\vec{p}$ is placed as shown then: 
The total number of turns and cross-section area in a solenoid is fixed. However, its length $L$ is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:
For a uniformly charged ring of radius $R,$ the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is:
Voltage rating of a parallel plate capacitor is $500 V.$ Its dielectric can withstand a maximum electric field of ${10}^{6}V/m.$ The plate area is ${10}^{-4} {m}^{2}$ . What is the dielectric constant if the capacitance is $15 pF$ ? $(given{\epsilon }_{0}=8.86\times {10}^{-12}{C}^{2}/N{m}^{2})$
The region between $y=0$ and $y=\mathrm{d}$ contains a magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B} \hat{\mathrm{z}}$. A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ enters the region with a velocity $\vec{v}=v \hat{i} .$ if $\mathrm{d}=\frac{\mathrm{m} v}{2 \mathrm{qB}},$ the acceleration of the charged particle at the point of its emergence at the other side is :
A solid metal cube of edge length $2 \mathrm{cm}$ is moving in the positive $y$-direction, at a constant speed of $6 m{s}^{-1}$. There is a uniform magnetic field of $\text{0.1} T$ in the positive $z$-direction. The potential difference between the two faces of the cube, perpendicular to the $x$-axis, is
A $10 m$ long horizontal wire extends from North East to South West. It is falling with a speed of $5.0 m{s}^{-1},$ at right angles to the horizontal component of the earth's magnetic field of $0.3\times {10}^{-4}\mathrm{Wb}{m}^{-2}.$ The value of the induced emf in the wire is:
A galvanometer, whose resistance is $50 \mathrm{ohm}$ , has $25$ divisions in it. When a current of $4\times {10}^{-4}$ A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5 V$ it should be connected to a resistance of:
Three charges $Q,+y$ and $+q$ are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $\mathrm{Q}$ is 
The mean intensity of radiation on the surface of the Sun is about ${10}^{8}W/{m}^{2}.$ The rms value of the corresponding magnetic field is closest to: