A proton moving with a velocity, $2.5 \times {10^7}\,m/s$, enters a magnetic field of intensity $2.5\,T$ making an angle ${30^o}$ with the magnetic field. The force on the proton is

A thin hollow copper pipe carries direct current than which is incorrect

An electron moves with a speed of $2 \times 10^5\, m/s$ along the $+ x$ direction in a magnetic field $\vec B = \left( {\hat i - 4\hat j - 3\hat k} \right)\,tesla$. The magnitude of the force (in newton) experienced by the electron is (the charge on electron $= 1.6 \times 10^{-19}\, C$)

If an electron is going in the direction of magnetic field $\overrightarrow B $ with the velocity of $\overrightarrow {v\,} $ then the force on electron is

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

An electron beam passes through a magnetic field of $2 \times 10^{-3}\,Wb/m^2$ and an electric field of $1.0 \times 10^4\,V/m$ both acting simultaneously. The path of electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively