Two long parallel conductors $S_{1}$ and $S_{2}$ are separated by a distance $10 \,cm$ and carrying currents of $4\, A$ and $2 \,A$ respectively. The conductors are placed along $x$-axis in $X - Y$ plane. There is a point $P$ located between the conductors (as shown in figure).
A charge particle of $3 \pi$ coulomb is passing through the point $P$ with velocity
$\overrightarrow{ v }=(2 \hat{ i }+3 \hat{ j }) \,m / s$; where $\hat{i}$ and $\hat{j} \quad$ represents unit vector along $x$ and $y$ axis respectively.
The force acting on the charge particle is $4 \pi \times 10^{-5}(-x \hat{i}+2 \hat{j}) \,N$. The value of $x$ is
$2$
$1$
$3$
$-3$
A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is ....
In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$
A charged particle of charge $q$ and mass $m$, gets deflected through an angle $\theta$ upon passing through a square region of side $a$, which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?
An electron is travelling horizontally towards east. A magnetic field in vertically downward direction exerts a force on the electron along
A particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{ i }(v \neq 0)$ towards a large screen placed in the $Y - Z$ plane at a distance $d.$ If there is a magnetic field $\overrightarrow{ B }= B _{0} \hat{ k },$ the minimum value of $v$ for which the particle will not hit the screen is