A particle having a mass of 10^{- 2} ,kg carries a charge of 5 × 10^{-8}, C. particle is given an i

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4.Moving Charges and Magnetism

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A particle having a mass of $10^{- 2} \,kg$ carries a charge of $5 \times 10^{-8}\, C.$ The particle is given an initial horizontal velocity of $10^5\, m/s $ in the presence of electric field $E$ and magnetic field $B.$ To keep the particle moving in a horizontal direction, it is necessary that

$(1)$ $\vec B$ should be perpendicular to the direction of velocity and $\vec E$ should be along the direction of velocity

$(2)$ Both $\vec B$ and $\vec E$ should be along the direction of velocity

$(3)$ Both $\vec B$ and $\vec E$ are mutually perpendicular and perpendicular to the direction of velocity.

$(4)$ $\vec B$ should be along the direction of velocity and $\vec E$ should be perpendicular to the direction of velocity

Which one of the following pairs of statements is possible?

$1$ and $ 3$

$3 $ and $ 4$

$2 $ and $ 3$

$2$ and $4$

(AIPMT-2010)

for statement $(2)$ :

If both the fields are along the direction of velocity of charged particle, magnetic field will not deflect the charged particle

and charged particle will move in the direction of electric field

for statement $(3) $:-

let us assume the charged particle is moving along the direction of $x-$axis of cartesian coordinate system.

If magnetic field is along $y-$axis, $v\times B$ force is acting along the direction $z-$axis . If electric fileld is in the direction of $z-$ axis,

then force due to electric field is exactly in opposite direction to the force due to electric field. By adjusting the field intensity,

we can get two forces acting on charged particle that are equal in magnitude but opposite in direction so that net force is zero

and charged particle will not be deflected

Standard 12

Physics

easy

easy

medium

easy

hard

(JEE MAIN-2024)