A charge particle of charge q- mass m is projected with a velocity v-v - The electric field E-E k- and magnetic field B -B - is applied- The acceleration of the particle isA- qvB - kB- qE E - m k-C- q E - vB k- mD- q E - vB k- m

The correct option is C q(E + vB)k̂mAcceleration = forcemass = Fm And we know that F_E = qE F_B = q(V×B) = q(V î× B ĵ) = q VB k̂ a = q(vB +E)mk̂ ...

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Byju's Answer

Standard XII

Physics

The Circular Motion

A charge part...

Question

A charge particle of charge q, mass m is projected with a velocity $\to v = v^i$ . The electric field $\to E = E^k$ and magnetic field $\to B = B^j$ is applied. The acceleration of the particle is

\frac{q v B}{m}^k

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\frac{q E}{m}^k

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\frac{q (E + v B)^k}{m}

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\frac{q (E - v B)^k}{m}

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Solution

The correct option is C $\frac{q (E + v B)^k}{m}$
Acceleration = $\frac{force}{mass} = \frac{F}{m}$
And we know that $F_{E} = q E$
$_{B} = q (\to V \times \to B) = q (V^i \times B^j) = q V B^k$
$a = \frac{q (v B + E)}{m}^k$

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A charge particle of charge q, mass m is projected with a velocity →v=v^i . The electric field →E=E^k and magnetic field →B=B^j is applied. The acceleration of the particle is

The correct option is C q(E+vB)^kmAcceleration = forcemass=Fm And we know that FE=qE −→FB=q(→V×→B)=q(V^i×B^j)=qVB^k a=q(vB+E)m^k

A charge particle of charge q, mass m is projected with a velocity $\to v = v^i$ . The electric field $\to E = E^k$ and magnetic field $\to B = B^j$ is applied. The acceleration of the particle is

The correct option is C $\frac{q (E + v B)^k}{m}$
Acceleration = $\frac{force}{mass} = \frac{F}{m}$
And we know that $F_{E} = q E$
$_{B} = q (\to V \times \to B) = q (V^i \times B^j) = q V B^k$
$a = \frac{q (v B + E)}{m}^k$