Solve Study Textbooks Guides

Use app

Login

>>Class 12
>>Physics
>>Moving Charges and Magnetism
>>Motion of Charged Particle Combined in Magnetic Field and Electric Fields
>> $A charged particle with charge q enters$

Question

A charged particle with charge $q$ enters a region of constant uniform and mutually orthogonal fields $E$ and $B$ with a velocity $v$ perpendicular to both $E$ and $B$ and comes out without any change in magnitude and direction of $v$ . Then,

A

$v = E \times B / B^{2}$

B

$v = E \times E / B^{2}$

C

$v = E \times E / E^{2}$

D

$v = B \times E / E^{2}$

Easy

Open in App

Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is A)

Since there is no change in magnitude and direction of particle velocity. So, Force due to magnetic field is equal to force due to electric field.
$\Rightarrow q (V \times B) = - q E$
$\Rightarrow V \times B = - E$
$\Rightarrow V = E \times B / B^{2}$

Solve any question of Moving Charges and Magnetism with:-

Patterns of problems

Was this answer helpful?

0

0

Similar questions

In a certain region of space electric field $E$ and magnetic field $B$ are perpendicular to each other. An electron enters in the region perpendicular to the direction of $B$ and $E$ both. If it moves un-deflected, then velocity of the electron is:

Let $E$ and $B$ denote electric and magnetic fields in a frame S and $E^{'}$ and $B^{'}$ in another frame S' moving with respect to S at a velocity $v$ . Two of the following equations are wrong. Identify them:

This question has multiple correct options

A charged particle is projected in a magnetic field $B = (3 i + 4 j) \times 1 0^{- 2} T$ and the acceleration is found to be $a = (x i + 2 j) m / s^{2}$ . The value of x is :

A proton is fired from origin with velocity $v = v_{0} j + v_{0} k$ in a uniform magnetic field $B = B_{0} j$ . In the subsequent motion of the proton:

This question has multiple correct options

A charged particle having charge q experiences a force $F = q (- j + k) N$ in a magnetic field B when it has a velocity $v_{1} = 1 \hat{i} m / s$ . The force becomes $F = q (i - k) N$ when the velocity is changed to $v_{2} = 1 \hat{j} m / s e c$ . The magnetic induction vector at that point is :

More From Chapter

Moving Charges and Magnetism

View chapter >

Revise with Concepts

Motion in the Presence of Electric and Magnetic Field

ExampleDefinitionsFormulaes

Cyclotron

ExampleDefinitionsFormulaes

Learn with Videos

Force on a moving charge in magnetic and electric fields

Construction, Working and Application of a Cyclotron - I

Construction, Working and Application of a Cyclotron - II

Practice more questions

Medium Questions

Shortcuts & Tips

Important Diagrams

Common Misconceptions

Problem solving tips

Memorization tricks