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Question

(D) The Lorentz force on a charge $q$ is given by $F = q(E + v 	imes B)$.For the particle to move with a constant velocity, the net force must be zer

Vedclass · Accepted Answer

$v 	imes E = B$

Vedclass · Answer

(D) The Lorentz force on a charge $q$ is given by $F = q(E + v 	imes B)$.For the particle to move with a constant velocity, the net force must be zero, so $F = 0$.This implies $E + (v 	imes B) = 0$, or $E = -(v 	imes B) = B 	imes v$.Since $E$ is equal to the cross product of $v$ and $B$, $E$ must be perpendicular to both $v$ and $B$.Therefore, $E \cdot v = 0$ and $E \cdot B = 0$. Thus, options $(a)$ and $(b)$ are true.If $v \cdot B = 0$, then $v$ is perpendicular to $B$. Given $E = -(v 	imes B)$, we can take the cross product with $B$:$E 	imes B = -(v 	imes B) 	imes B = -[(v \cdot B)B - (B \cdot B)v]$.Since $v \cdot B = 0$, this simplifies to $E 	imes B = (B \cdot B)v$, which gives $v = \frac{E 	imes B}{B \cdot B}$. Thus, option $(c)$ is true.Finally, $v 	imes E = v 	imes (-(v 	imes B)) = -(v 	imes (v 	imes B)) = -[(v \cdot B)v - (v \cdot v)B]$.This is generally not equal to $B$. Therefore, option $(d)$ is not true.

$A$ charge $q$ enters a region having electric field $E$ and magnetic field $B$ with velocity $v$. If it continues to move with the same velocity, then which of the following statements is not true?

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