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An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

$A$ point charge is placed in a moving train. $A$ passenger $A$ sitting in the train and a person $B$ standing on the ground observe the fields due to this charge. Then:

In a region,there are uniform and constant electric and magnetic fields. Both these fields are parallel to each other. $A$ stationary charged particle is released in this region. The path of the particle will be.......

$A$ positively charged particle $q$ of mass $m$ is passed through a velocity selector. It moves horizontally rightward without deviation along the line $y = \frac{2mv}{qB}$ with a speed $v$. The electric field is vertically downwards and the magnetic field is into the plane of the paper. Now,the electric field is switched off at $t = 0$. The angular momentum of the charged particle about the origin $O$ at $t = \frac{\pi m}{qB}$ is:

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