$OABC$ is a current carrying square loop an electron is projected from the centre of loop along its diagonal $AC$ as shown. Unit vector in the direction of initial acceleration will be
$\hat k$
$ - \,\left( {\frac{{\hat i\, + \,\hat j}}{{\sqrt 2 }}} \right)$
$=- \hat k$
$\frac{{\hat i\, + \,\hat j}}{{\sqrt 2 }}$
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 initial direction of motion (see figure), the value of $sin\,\alpha $ will be
charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then
If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be
The electron in the beam of a television tube move horizontally from south to north. The vertical component of the earth's magnetic field points down. The electron is deflected towards
Two parallel wires in the plane of the paper are distance $X _0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have direction opposite to each other, the radius of curvature of the path is $R_2$.
If $\frac{x_0}{x_1}=3$, the value of $\frac{R_1}{R_2}$ is.