A particle of mass m and charge q is thrown from origin at t-0 with velocity 2-3-4k- units in a region with uniform magnetic field B-2- units- After time t- mqB- an electric field E- is switched on- such that particle moves on a straight line with constant speed- E- may be

The correct option is A 6k̂+8ĵ unitsIn T2 , v_⊥ gets reversed v⃗=2î 3ĵ 4k̂ E⃗+v⃗×B⃗=0 E⃗= (v⃗×B⃗) = (2î 3ĵ 4k̂)× 2î = ...

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Standard XII

Physics

Charging by Friction

A particle of...

Question

A particle of mass m and charge q is thrown from origin at $t = 0$ with velocity $2^i + 3^j + 4^k$ units in a region with uniform magnetic field $\to B = 2^i$ units. After time $t = \frac{π m}{q B}$ , an electric field $\to E$ is switched on, such that particle moves on a straight line with constant speed. $\to E$ may be

5^k - 10^j

units

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- 6^k - 9^j

units

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- 6^k + 8^j

units

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6^k + 8^j

units

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Solution

The correct option is A $- 6^k + 8^j$ units
In $\frac{T}{2}$ ,
$v_{⊥}$ gets reversed
$\to v = 2^i - 3^j - 4^k$
$\to E + \to v \times \to B = 0$
$\to E = - (\to v \times \to B)$
$= - (2^i - 3^j - 4^k) \times 2^i$
$= - 6^k + 8^j$

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A particle of mass m and charge q is thrown from origin at t=0 with velocity 2^i+3^j+4^k units in a region with uniform magnetic field →B=2^i units. After time t=πmqB, an electric field →E is switched on, such that particle moves on a straight line with constant speed. →E may be

The correct option is A −6^k+8^j unitsIn T2,v⊥ gets reversed→v=2^i−3^j−4^k→E+→v×→B=0→E=−(→v×→B)=−(2^i−3^j−4^k)×2^i=−6^k+8^j

The correct option is A $- 6^k + 8^j$ units
In $\frac{T}{2}$ ,
$v_{⊥}$ gets reversed
$\to v = 2^i - 3^j - 4^k$
$\to E + \to v \times \to B = 0$
$\to E = - (\to v \times \to B)$
$= - (2^i - 3^j - 4^k) \times 2^i$
$= - 6^k + 8^j$