A particle of mass m and charge q has an initial velocity v = v₀j. If an electric field E = E₀i and magnetic field B = B₀i act on the particle, its speed will double after a time: (a) (3mv₀)/(qE₀) (b) (√(2)mv₀)/(qE₀) (c) (√(3)mv₀)/(qE₀) (d) (2mv₀)/(qE₀)

Answer: (c). B is along x and v₀ ⊥ B, so the yz motion is circular with constant speed v₀. Along x only E acts: vx = qE₀/mt. Speed = √(v₀² + vx²) = 2v₀ ⇒ vx = √(3)v₀. t = (√(3)mv₀)/(qE₀). So the answer is (c).

A particle of mass m and charge q has an initial velocity v = v₀j. If… | JEE Main 2020 (7 Jan, Shift 2)

Q84JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Moving Charge in a Magnetic Field

A particle of mass m and charge q has an initial velocity $\vec{v} = v_0\hat{j}$ . If an electric field $\vec{E} = E_0\hat{i}$ and magnetic field $\vec{B} = B_0\hat{i}$ act on the particle, its speed will double after a time:

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Answer: (c)

$\vec{B}$ is along $x$ and $\vec{v}_0 \perp \vec{B}$ , so the $yz$ motion is circular with constant speed $v_0$ .
Along $x$ only $\vec{E}$ acts: $v_x = \dfrac{qE_0}{m}t$ .
Speed $= \sqrt{v_0^2 + v_x^2} = 2v_0 \Rightarrow v_x = \sqrt{3}v_0$ .
$t = \dfrac{\sqrt{3}mv_0}{qE_0}$ . So the answer is (c).

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Moving Charge in a Magnetic Field question from JEE Main 2020 (7 Jan, Shift 2)

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