Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
An electron with energy $0.1\,ke\,V$ moves at right angle to the earth's magnetic field of $1 \times 10^{-4}\,Wbm ^{-2}$. The frequency of revolution of the electron will be. (Take mass of electron $=9.0 \times 10^{-31}\,kg$ )
A straight wire of length $0.5\, metre$ and carrying a current of $1.2\, ampere$ placed in a uniform magnetic field of induction $2\, Tesla$. The magnetic field is perpendicular to the length of the wire. The force on the wire is.......$N$
An element $\Delta l=\Delta \mathrm{xi}$ is placed at the origin and carries a large current $\mathrm{I}=10 \mathrm{~A}$. The magnetic field on the $y$-axis at a distance of $0.5 \mathrm{~m}$ from the elements $\Delta \mathrm{x}$ of $1 \mathrm{~cm}$ length is:
A particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then
A thin circular wire carrying a current $I$ has a magnetic moment $M$. The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment
A current of $0.1\, A$ circulates around a coil of $100$ $turns$ and having a radius equal to $5\, cm$. The magnetic field set up at the centre of the coil is $({\mu _0} = 4\pi \times {10^{ - 7}}\,weber/ampere - metre)$