A particle having a mass of 10 - 2 kg carries a charge of 5 × 10 - 8 C. The particle is given an initial horizontal velocity of 105 m s ⁻¹ 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) B should be perpendicular to the direction of velocity and E should be along the direction of velocity (2) Both B and E should be along the direction of velocity (3) Both B and E are mutually perpendicular and perpendicular to the direction of velocity. (4) B should be along the direction of velocity and E should be perpendicular to the direction of velocity Which one of the following pairs of statements is possible? (a) (1) and (3) (b) (3) and (4) (c) (2) and (3) (d) (2) and (4)

Answer: (c). (1) B ⊥ v gives a sideways force that E along v cannot cancel: not possible. (2) B ∥ v gives no magnetic force; E ∥ v only changes speed, so the motion stays horizontal: possible. (3) Crossed E and B, both ⊥ v, can cancel ( qE = qvB, velocity selector): possible. (4) B ∥ v gives no magnetic force, so E ⊥ v deflects the particle: not possible. So the answer is (c) (2) and (3).

A particle having a mass of 10 - 2 kg carries a charge of 5 × 10 - 8… | NEET 2010 (AIPMT, Mains)

Q108NEET 2010 · AIPMT · MAINSMedium

Moving Charge in a Magnetic Field

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 105 m s ${}^{-1}$ in the presence of electric field $\vec{E}$ and magnetic field $\vec{B}$ . 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?

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Answer: (c)

(1) $\vec{B} \perp \vec{v}$ gives a sideways force that $\vec{E}$ along $\vec{v}$ cannot cancel: not possible.
(2) $\vec{B} \parallel \vec{v}$ gives no magnetic force; $\vec{E} \parallel \vec{v}$ only changes speed, so the motion stays horizontal: possible.
(3) Crossed $\vec{E}$ and $\vec{B}$ , both $\perp \vec{v}$ , can cancel ( $qE = qvB$ , velocity selector): possible.
(4) $\vec{B} \parallel \vec{v}$ gives no magnetic force, so $\vec{E} \perp \vec{v}$ deflects the particle: not possible.
So the answer is (c) (2) and (3).

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