Question

In a certain region, static electric and magnetic fields exist. The magnetic field is given by $\vec{B} = B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$. If a test charge given by $\vec{v} = v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)$ experiences no force in that region, then the electric field in that region, in SI units is:
A. $\vec{E} = -v_0B_0\left(\hat{i} +\hat{j}-7\hat{k}\right)$
B. $\vec{E} = -v_0B_0\left(14\hat{j}+7\hat{k}\right)$
C. $\vec{E} = v_0B_0\left(14\hat{j}+7\hat{k}\right)$
D. $\vec{E} = -v_0B_0\left(3\hat{i} -2\hat{j}-4\hat{k}\right)$

Answer 1

Hint: Both the electric field and the magnetic field exert a force on the test charge and influence its trajectory. In such a case, arrive at the resultant expression for the net force acting on the test charge, which we know to be the Lorentz force. Recall that if a test charge experiences no force, it traverses through the electromagnetic field at a constant velocity. To this end, calculate the appropriate magnitude of the existing electric field .

Formula used:
Lorentz force: $\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right)$

Complete step-by-step answer:
We know that electric and magnetic fields are invisible fields of force that are responsible for influencing charged particles that traverse through them and define their trajectory. The electric and magnetic forces thus exerted on the particle have a resultant effect on the particle’s motion through the electromagnetic space.
This resultant force acting on a charged particle moving under the influence of the electric and magnetic forces is known as the Lorentz force. If a particle of charge$\;q$ moving with a velocity $\vec{v}$ in and electric field $\vec{E}$ and magnetic field $\vec{B}$, the Lorentz force $\vec{F}$ experienced by it is given as:
$\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right)$
In the context of the question, we are given that:
$\vec{B} = B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$ and $\vec{v} = v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)$
 And the charged particle experiences no force in the electromagnetic region, i.e.,
$\vec{F} = 0$
$\Rightarrow q\left(\vec{E} + (\vec{v} \times \vec{B})\right) = 0$
$\Rightarrow \vec{E} + (\vec{v} \times \vec{B}) = 0$
$\Rightarrow \vec{E} = -(\vec{v} \times \vec{B})$
Plugging in the given values we get:
$\vec{E} = - \left(v_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right)\right) \times \left(B_0 \left(\hat{i} + 2\hat{j}-4\hat{k}\right)\right)$
$\Rightarrow \vec{E} = -v_0B_0 \left(3\hat{i} +\hat{j}-2\hat{k}\right) \times \left(\hat{i} + 2\hat{j}-4\hat{k}\right)$
$\Rightarrow \vec{E} = -v_0B_0\left(-\hat{k} -2\hat{j} -6\hat{k}+4\hat{i}-12\hat{j}-4\hat{i}\right) = -v_0B_0\left(+4\hat{i}-4\hat{i}-2\hat{j}-12\hat{j}-\hat{k}-6\hat{k}\right)$
$\Rightarrow \vec{E} = -v_0B_0 \left(0\hat{i}-14\hat{j}-7\hat{k}\right) = v_0B_0\left(14\hat{j} + 7\hat{k}\right)$

So, the correct answer is “Option C”.

Note: A common misconception is that in the Lorentz force expression, the velocity or trajectory of the charged particle and the magnetic field that it traverses through must be perpendicular to each other for a finite magnetic force to act on it. This, however, is not the case, since the cross product between$\vec{v}$ and $\vec{B}$ suggests that:
$\vec{F} = q\left(\vec{E} + (\vec{v} \times \vec{B})\right) = q\left(\vec{E} + (|\vec{v}|| \vec{B}|\;sin\theta)\right)$, where $\theta$ can be any angle between $\vec{v}$ and $\vec{B}$.
Additionally, the Lorentz force acting on the particle will always be perpendicular to the plane containing $\vec{v}$ and $\vec{B}$.