A proton is projected with a velocity 10^7, m/s , at right angles to a uniform magnetic field of ind

English

Gujarati

Hindi

4.Moving Charges and Magnetism

medium

A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$

A

$0.81\times10^{-7}$

B

$1.62\times10^{-7}$

C

$2.43\times10^{-7}$

D

$3.24\times10^{-7}$

Solution

Solution is Lorem ipsum dolor sit amet, consectetur adipiscing elit. Ut elit tellus, luctus nec ullamcorper mattis, pulvinar dapibus leo.

Standard 12

Physics

Share

Similar Questions

Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :…….$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )

medium

(JEE MAIN-2020)

Give features of force on charge particle inside magnetic field.

medium

An electron moves with speed $2 \times {10^5}\,m/s$ along the positive $x$-direction in the presence of a magnetic induction $B = \hat i + 4\hat j – 3\hat k$ (in $Tesla$) The magnitude of the force experienced by the electron in Newton's is (charge on the electron =$1.6 \times {10^{ – 19}}C)$

medium

The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat z$ A particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{{mv}}{{2qB}}$ , the acceleration of the charged particle at the point of its emergence at the other side is

hard

(JEE MAIN-2019)

A charged particle of mass $\mathrm{m}$ and charge $q$ moving under the influence of uniform electric field $E\hat{i }$ and a uniform magnetic field $B\hat{k}$ follows a trajectory from point $\mathrm{P}$ to $\mathrm{Q}$ as shown in figure. The velocities at $P$ and $Q$ are respectively, $v\hat i$ and $-2 v \hat j$. Then which of the following statements $(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})$ are the correct $?$ (Trajectory shown is schematic and not to scale)
$(A)$ $\mathrm{E}=\frac{3}{4}\left(\frac{\mathrm{mv}^{2}}{\mathrm{qa}}\right)$
$(B)$ Rate of work done by the electric field at $\mathrm{P}$ is $\frac{3}{4}\left(\frac{\mathrm{mv}^{3}}{\mathrm{a}}\right)$
$(C)$ Rate of work done by both the fields at $\mathrm{Q}$ is zero
$(D)$ The difference between the magnitude of angular momentum of the particle at $\mathrm{P}$ and $Q$ is $2 mav$.

hard

(JEE MAIN-2020)