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The force on a charged particle due to electric and magnetic fields is given by $\vec{F}=q\vec{E}+q\vec{v}\times\vec{B}$. If $\vec{E}$ is along the $X$-axis and $\vec{B}$ asking the $Y$-axis, in what direction and with what minimum speed $v$ should a positively charged particle be sent so that the best force on it is $\mathrm{ZERO}$?

I have done it like: For force to be $0$, $||q\vec{v}\times\vec{B}||=0$ and has to act opposite to the direction of $q\vec{E}$ and so $||\vec{v}||=\frac{-||\vec{E}||}{||\vec{B}||\sin\theta}$.

I need to know certain things.

  1. I understand that the velocity has to be sent in the $Z$-axis but whether it's positive $Z$ axis or negative $Z$ axis will depend on the direction of the electric and magnetic fields right?

  2. So what should be the speed: $\frac{||\vec{E}||}{||\vec{B}||}$ right? And the velocity should be $\frac{±||\vec{E}||}{||\vec{B}||}$ along $Z$ axis right? (Depending upon positive or negative $Z$ axis?)

  3. By any means, do we know the direction (positive or negative axis) of the velocity, electric field and magnetic field from what's mentioned in the question?

Please check my solution and if possible, do write a fair solution for me as well in the answer section.

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  • $\begingroup$ "along the X axes" probably means here in the positive direction of x. Same for y. $\endgroup$ – Ofek Gillon May 14 '17 at 3:39
  • $\begingroup$ Because like you said, if not, we don't have enough details to solve the question. Sometimes when people say "along the axes" they mean parallel to it and specifically not anti parallel. This interpretation of the question is vital for getting a solution :) $\endgroup$ – Ofek Gillon May 14 '17 at 3:40
  • $\begingroup$ Yeah, true. So my reasoning was correct, right? But the fact that charge is sent in the positive or negative axis has no relation to it being positive or perhaps negative, right? I know the total force will depend upon it being negative or positive but the positivity/negativity of the charge won't be represented by the positive/negative axis I suppose. So how would it be represented? By using (+,-) sign. Am I correct? $\endgroup$ – Mathejunior May 14 '17 at 3:45
  • $\begingroup$ In the end of the question, the charge of the particle is given. $\endgroup$ – Mitchell May 14 '17 at 3:47
  • $\begingroup$ @BhavyaSharma Yeah, but that doesn't mean if it's in the positive or negative axis right? $\endgroup$ – Mathejunior May 14 '17 at 3:54
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When the direction of the vectors is not specified, 'along the axes' by default means along the +ve axis.

The charged particle when sent along $-z$ $axis$, will experience a force ,due to magnetic field and the electric field, towards $+x$ $axis$. In this case the net force cannot be zero whatever the speed of the charged particle may be.

The net force can only be $zero$ if the charged particle moves along $+z$ $axis$.

and specifically, the angle between the velocity and the magnetic field has to be $90^{\circ}$, only then the magnetic force can totally cancel out the electric force. For two forces to cancel each other, they have to be anti-parallel and on the same line.

Therefore,

$\vec{E}=-\vec{v} \times \vec{B}$

Here the -ve sign means only one thing, the vector that we get from the cross product of $\vec{v}$ and $\vec{B}$ is opposite to the electric field and equal in magnitude.

The direction of the velocity can also be interpreted from this relation.

$\vec{E}= +\hat{i}$

$\vec{B}=+\hat{j}$

For, $\vec{v} \times \vec{B} = -\hat{i} $, $\vec{v}$ has to be directed along $+\hat{k}$,

because, $\hat{k}\times \hat{j}=-\hat{i}$

On comparing direction of right hand side and left hand side,

$\hat{i}=-(-\hat{i})$

LHS = RHS

$|\vec{v}|=\frac{|\vec{E}|}{|\vec{B}|}$ , This is certain.

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