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# a particle having charge q and mass m is projected with velocity

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## A charged particle having charge q and mass m is projected into a region of uniform electric field of strength E0, with velocity V0 perpendicular to E0. Throughout the motion, apart from electric force, the particle also experiences a dissipative force of constant magnitude qE 0 and directed opposite to its velocity. If V 0

A charged particle having charge q and mass m is projected into a region of uniform electric field of strength E0, with velocity V0 perpendicular to E0. Throughout the motion, apart from electric force, the particle also experiences a dissipative force of constant magnitude qE 0 and directed opposite to its velocity. If V 0|=6 m / s, then find its speed when it has turned through an angle of 90∘.Write speed in m / s.

Coefficients of Friction

A charged par... Question

A charged particle having charge

q and mass m

is projected into a region of uniform electric field of strength

−→ E 0 , with velocity → V 0 perpendicular to −→ E 0

. Throughout the motion, apart from electric force, the particle also experiences a dissipative force of constant magnitude

q E 0

and directed opposite to its velocity. If

→ V 0 | = 6 m/s

, then find its speed when it has turned through an angle of

90 ∘ . Write speed in m/s. Open in App Solution

Here v

is speed which is equal to

v = √ v 2 x + v 2 y and d v d t =

rate of change in speed

Consider the situation when the particle has turned through an angle of

ψ d v d t = − q E 0 m [ 1 − sin ψ ] d v x d t = q E 0 m [ 1 − sin ψ ] ⇒ d v d t = − d v x d t ⇒ v = − v x + C At t = 0 , v = v 0 and v x = 0 ⇒ C = v 0 ⇒ v = v 0 − v x = v 0 − v sin ϕ

When it has turned through an angle of

90 ∘ , v = v 0 1 + sin ϕ = 6 1 + 1 = 3 m/s Suggest Corrections 1

SIMILAR QUESTIONS

Q. In a certain region free from gravity, electric field is along negative x-direction and it is constant. A particle having mass m and charge q is projected along x-direction with speed

V 0

→ F = → C × → V

is acting on the charge where

→ V

is velocity vector and

→ C

is a constant vector. The charge comes out of region with speed

V 0 2

as shown in figure, then the magnitude of electric field is :

Q. A charged particle is projected perpendicular to a uniform electric field of length

2 m

and initial velocity of charged particle is

10 m/s

. Calculate deflection of the charge in the direction of electric field., when it leaves it.

Q. A charged particle

( mass m and charge q ) moves along x − axis with velocity v 0

. When it passes through the origin, it enters a region having uniform electric field

E = − E ^ j which extends up to x = d

. The equation of path of electron in the region

x > d is :

Q. A particle of charge

q and mass m

starts moving from the origin under the action of electric field

→ E = E 0 ^ i and → B = B 0 ^ i with velocity → V = V 0 ^ j

. The speed of particle will become

2 V 0 after a time:

Q. A charged particle is projected perpendicular to a uniform electric field of length

2 m

and initial velocity of charged particle is

10 m/s

. Calculate deflection of the charge in the direction of electric field., when it leaves it.

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## A particle having charge q and mass m is projected from the origin with v..

Solution For A particle having charge q and mass m is projected from the origin with velocity v0​[i^+k^] in a uniform magnetic field B=−B0​k at time t=0. If x(t),y(t),z(t) represent x,y and z-co-ordin

Home Class 12 Physics

Moving Charges and Magnetism

Motion in a Magnetic Field

Question

Solving time: 5 mins

A particle having charge q and mass m is projected from the origin with velocity v

0 ​ [ i ^ + k ^

] in a uniform magnetic field

B =−B 0 ​ k

at time t=0. If x(t),y(t),z(t) represent x,y and z-co-ordinates of the particle at time t, then select the correct option(s).

SELECT MULTIPLE OPTION

A y(t)= qB mv 0 ​ ​ [1−cos( m qB ​ t)] B x(t)= qB mv 0 ​ ​ cos( m qBt ​ ) C y(t)= qB mv 0 ​ ​ [1−sin( m qBt ​ )] D z(t)=v 0 ​ t

Difficulty level: Hard

Viewed by: 5,912 students

## Solutions

(1)

R= qB mv 0 ​ ​ ,ω= R v 0 ​ ​ = m qB ​ y(t)=R−Rcosωt y(t)= qB mV 0 ​ ​ (1−cos m qBt ​ ) and Z(t)=V 0 ​ t 89 Share

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## A particle of charge q and mass m is projected with a velocity v _0 towards a circular region having a uniform magnetic field B perpendicular and into the plane of paper from point P as shown in the figure. R is the radius and O is the center of the circ

Answer to: A particle of charge q and mass m is projected with a velocity v _0 towards a circular region having a uniform magnetic field B...

Magnetic fields

## A particle of charge q and mass m is projected with a velocity v _0 towards a circular region...

A particle of charge q and mass m is projected with a velocity v _0 towards a circular region... Question:

A particle of charge

q q and mass m m

is projected with a velocity

v 0 v0

towards a circular region having a uniform magnetic field

B B

perpendicular and into the plane of paper from point

P P

as shown in the figure.

R R is the radius and O O

is the center of the circular region. If the line

O P OP makes an angle θ θ

with the direction of

v 0 v0

, then determine the value of

v 0 v0

so that particle passes through

O O .

## Magnetic field

When a magnetic field is applied to a charged particle, a force is exerted on the charge, which causes it to move in an orbicular path with a certain velocity. That force will be a centripetal force in the magnetic field.

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Let r is the radius of the circular path.

Therefore, the angle can be calculated as,

{eq}\begin{align*} \sin \theta &=...

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The Effect of a Magnetic Field on Moving Charges: Physics Lab

from

Chapter 16 / Lesson 4

3.6K

A magnetic field is the area in which other objects can be affected by a magnet or moving charge. Observe the effect magnetic fields have on moving charges with this lab and analyze the observational data collected.

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Mohammed 13 day ago

Guys, does anyone know the answer?