A particle of mass 1- 10-26 kg and charge 1-6- 10-19 C travelling with a velocity 1-28- 106 m-s along the positive X-axis enters a region in which a uniform electric field E- and a uniform magnetic field of induction B- are present- if E-102-4 - 103 k- NC-1 and B-8 - 10-2 - Wbm-2- the direction of motion of the particles is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Properties of Field Lines

A particle of...

Question

A particle of mass $1 \times 10^{- 26} k g$ and charge $1.6 \times 10^{- 19} C$ travelling with a velocity $1.28 \times 10^{6} m / s$ along the positive $X$ -axis enters a region in which a uniform electric field $\to E$ and a uniform magnetic field of induction $\to B$ are present. if $\to E = - 102.4 \times 10^{3}^k N C^{- 1}$ and $\to B = 8 \times 10^{- 2}^j W b m^{- 2}$ , the direction of motion of the particles is

along the positive X-axis

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

along the negative X-axis

No worries! We‘ve got your back. Try BYJU‘S free classes today!

45^{O}

to the positive X-axis

No worries! We‘ve got your back. Try BYJU‘S free classes today!

135^{O}

to the postive X-axis

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B along the positive X-axis
Write Force
F = q(vXB) + qE.
F =0N.
Hence the particle will continue moving towards positive X-axis.

Suggest Corrections

Similar questions

Q. A particle of mass

1 \times 10^{- 26}

k g

and charge

1.6 \times 10^{- 19}

C

travelling with a velocity

1.28 \times 10^{6} m s^{- 1}

along the positive

X -

axis enters a region in which there is a uniform electric field

\to E = - 102.4 \times 10^{3} k N C^{- 1}

and magnetic field

\to B = 8 \times 10^{- 2}^j W b m^{- 2}

, the direction of motion of the particles is :

Q. A particle of mass

1 \times 10^{- 26} kg

and charge

1.6 \times 10^{- 19} C

travelling with a velocity

1.28 \times 10^{6} {ms}^{- 1}

along the positive

X

-axis enters a region in whch a uniform electric field

\to E = (- 102.4 \times 10^{3}^k) \frac{N}{C}

and magnetic field

\to B = (8 \times 10^{- 2})^j \frac{W b}{m^{2}}

exists. The direction of motion of the particle is along

Q. A particle of mass

1 \times 10^{- 26} k g

and charge

+ 1.6 \times 10^{- 19} C

travelling with a velocity

1.28 \times 10^{6} m s^{- 1}

in the +x direction enters a region in which uniform electric field E and a uniform magnetic field of induction B are present such that

E_{x} = E_{y} = 0, E_{z} = - 102.4 k V m^{- 1}

, and

B_{x} = B_{z} = 0, B_{y} = 8 \times 10^{- 2}

. The particle enters this region at time

t = 0

. Determine the location (x, y, z coordinates) of the particle as

t = 5 \times 10^{- 6} s

. If the electric field is switched off at this instant (with the magnetic field present), what will be the position of the particle at

t = 7.45 \times 10^{- 6} s

Q. Assertion :A charged particle moves along positive y-axis with constant velocity in uniform electric and magnetic fields.If magnetic field is acting along positive x-axis,then electric field should act along positive z-axis. Reason: To keep the charged particle undeviated the relation

\to E = \to B \times \to v

must hold good.

Q. A particle of charge

q = 16 \times 10^{- 18}

C moving with

10 m s^{- 1}

along

x -

axis enter a magnetic field of induction

B

along the

y -

axis and an electric field

10^{4}

^{- 1}

along negative

z -

direction. If the particle continues to move along

x -

axis then the strength of magnetic field is

Join BYJU'S Learning Program

The correct option is B along the positive X-axisWrite ForceF = q(vXB) + qE.F =0N.Hence the particle will continue moving towards positive X-axis.

The correct option is B along the positive X-axis
Write Force
F = q(vXB) + qE.
F =0N.
Hence the particle will continue moving towards positive X-axis.