Correct Answer is Option (C)

Correct Answer is Option (A)

$B=\frac{{\mu }_{0}I}{2\pi R}$

F = BVq sin$\theta$

$\theta$ = 90${}^{\circ }$

F = BVq

$F=\frac{{\mu }_{0}I}{2\pi R}\times V\times e=\frac{2\times {10}^{-7}\times 5}{20\times {10}^{-2}}\times {10}^5\times 1.6\times {10}^{-19}$

F = 8 $\times$ 10^$-$20 N

Correct Answer is Option (D)

Radius of the path (r) = $\frac{mv}{qB}$ = $\frac{p}{qB}$

So, the radius for H-atom r_H = $\frac{p}{eB}$

The radius for $\alpha$ particle r_$\alpha$ = $\frac{p}{2eB}$

$\therefore$ $\frac{r_H}{r_{\alpha }}=\frac{\frac{p}{eB}}{\frac{p}{2eB}}$ = $\frac{2}{1}$

Correct Answer is Option (A)

Frequency of revolution of charge in magnetic field is given as

F = $\frac{qB}{2\pi m}$

= $\frac{\left(1.76\times {10}^{11}\right)\times 3.57\times {10}^{-2}}{2\times 3.14}$

= 10⁹ Hz

= 1 GHz

Correct Answer is Option (B)

As we know, F = qvB = $\frac{m{v}^2}{R}$

$\therefore$ $R=\frac{mv}{qB}=\frac{\sqrt{2m\left(kE\right)}}{qB}$

Since R is same so, $KE\propto \frac{q^2}{m}$

Therefore KE of $\alpha$ particle

$=\frac{q^2}{m}=\frac{{\left(2\right)}^2}{4}=1MeV$

Correct Answer is Option (B)

The situation is as shown in the figure.


Here, v = 10⁷ m/s, B = 2 × 10^–4 Wb/m²

The magnitude of the force experienced by the electron is

F = evBsin$\theta$

   ($\because \overrightarrow{v}and\overrightarrow{B}are$ perpendicular to each other)

= evBsin90° = 1.6 × 10^–19 × 10⁷ × 2 × 10^–4 × 1

= 3.2 × 10^–16 N

Correct Answer is Option (D)

When moves with an acceleration a₀ towards west, electric field

$E=\frac{F}{q}=\frac{m{a}_0}{e}\left(West\right)$

When moves with an acceleration 3a₀ towards east, magnetic field

$B=\frac{2m{a}_0}{e{v}_0}\left(downward\right)$

Correct Answer is Option (B)

According to the principal of circular motion in a magnetic field

$F_c=F_m\Rightarrow \frac{m{v}^2}{R}=qVB$

$\Rightarrow R=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{2m.k}}{qB}$

$R_{\alpha }=\frac{\sqrt{2\left(4m\right)K^{\prime }}}{2qB}$

$\frac{R}{R_{\alpha }}=\sqrt{\frac{K}{K^{\prime }}}$

but $R=R_{\alpha }$ (given)

$\therefore$ K = K' = 1 MeV

Correct Answer is Option (C)

Time period of cyclotron is

$T=\frac{1}{v}=\frac{2\pi m}{eB};B=\frac{2\pi m}{e}v;R=\frac{mv}{eB}=\frac{p}{eB}$

$\Rightarrow P=eBR=e\times \frac{2\pi mv}{e}R=2\pi mvR$

$K.E.=\frac{p^2}{2m}=\frac{{\left(2\pi mvR\right)}^2}{2m}=2{\pi }^{2}m{v}^2{R}^2$

Correct Answer is Option (B)

$\overrightarrow{v}$ and $\overrightarrow{B}$ are in same direction so that magnetic force on electron becomes zero, only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.

Correct Answer is Option (C)

Force due to electric field acts along the direction of the electric field but force due to the magnetic field acts along a direction perpendicular to both the velocity of the charged particle and the magnetic field. Hence both statements (2) and (3) are true. In statement (2), magnetic force is zero, so, electric force will keep the particle continue to move in horizontal direction. In statement (3), both electric and magnetic forces will be opposite to each other. If their magnitudes will be equal then the particle will continue horizontal motion.