An electric field and magnetic field $1.8 \times 10^4 \mathrm{Vm}^{-1}$ and $6 \times 10^{-3} \mathrm{~T}$ respectively are applied simultaneously on an electron beam such that path of the beam remains undeviated, then the speed of the electron will be:
$1.5 \times 10^6 \mathrm{~ms}^{-1}$
$3 \times 10^7 \mathrm{~ms}^{-1}$
$3 \times 10^6 \mathrm{~ms}^{-1}$
$1.5 \times 10^7 \mathrm{~ms}^{-1}$
Two point charges $P=+25 \mu C$ and $Q=-16 \mu C$ are placed 5 cm apart. Find the position of the point at which the resultant electric field is zero:
1 cm from $Q$ and 4 cm from $P$ on the dipole axis
2.5 cm from $Q$ and 2.5 cm from $P$ on the dipole axis
20 cm from $Q$ and 25 cm from $P$ on the dipole axis
25 cm from $Q$ and 20 cm from $P$ on the dipole axis
The basic principle used behind the working of electron microscope is:
Using charged mirrors to achieve the desired magnification.
Wave nature of electrons.
Electrostatic field created by a beam of electrons.
Magnifying power of very thin aperture convex lenses.
When a current of 2.5 A passes through the primary coil of a transformer of 200 number turns, the magnetic flux linked with the secondary coil having 400 turns is $600 \times 10^{-6} \mathrm{~T} \mathrm{~m}^2$. Find the induced emf in the secondary coil, when the current in the primary coil increases at a rate of $0.2 \mathrm{As}^{-1}$
$1.92 \times 10^{-2} V$
$1.92 \times 10^{-4} V$
$0.92 \times 10^{-4} V$
$0.92 \times 10^{-2} V$
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