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Assuming that particle as a charge ,force acting on a charge $q$ is

$$ F = q \left( v \times B \right) \tag{1} \label{1} $$

So let say $q$ is along $+y$ axis and $E$ and $B$ (both the fields) along $+x$ axis. Fine use right hand screw rule (curl your right hand from direction of $v$ to $B$ and thumb points the direction of force) along $v$ to $B$ (that is $v \times B$) and we see see the thumb point along $-v_{e} Z$ .

Now we also can see that the force is perpendicular to velocity of particle.

Power, if you remember is $F \cdot v$ or $F v \cos \left( \theta \right)$. $\theta = 90^{o}$ so power $= 0$ and work also as power = work/time. So, the particle will perform circular motion.

Here's why 2nd case won't happen. The helical motion is when velocity and electric field makes an acute angle but our $\theta$ here is $90^{o}$. If it's less than $90$, motion goes helical.

Assuming that particle as a charge ,force acting on a charge $q$ is

$$ F = q \left( v \times B \right) \tag{1} \label{1} $$

So let say $q$ is along $+y$ axis and $E$ and $B$ (both the fields) along $+x$ axis. Fine use right hand screw rule (curl your right hand from direction of $v$ to $B$ and thumb points the direction of force) along $v$ to $B$ (that is $v \times B$) and we see see the thumb point along $- Z$ .

Now we also can see that the force is perpendicular to velocity of particle.

Power, if you remember is $F \cdot v$ or $F v \cos \left( \theta \right)$. $\theta = 90^{o}$ so power $= 0$ and work also as power = work/time. So, the particle will perform circular motion.

Here's why 2nd case won't happen. The helical motion is when velocity and electric field makes an acute angle but our $\theta$ here is $90^{o}$. If it's less than $90$, motion goes helical.

Assuming that particle as a charge ,force acting on a charge q is F=q(v x B)

So let say q is along +y axis and E and B (both the fields) along +x axis ,

fine use right hand screw rule (curl your right hand from direction of v to B and thumb points the direction of force) along v to B ( that is v cross B) and we see see the thumb point along -ve Z

Now we also can see that The force is perpendicular to velocity of particle

And power if you remember is F•v or Fv * Cos theta

Theta = 90 so power =0 and work also as power = work/time

  So the particle will perform circular motion

And here's why 2nd case won't happen , the helical motion is when velocity and electric field makes an acute angle but our theta here is 90, if it's less than 90 motion goes helical

Assuming that particle as a charge ,force acting on a charge $q$ is

$$ F = q \left( v \times B \right) \tag{1} \label{1} $$

So let say $q$ is along $+y$ axis and $E$ and $B$ (both the fields) along $+x$ axis. Fine use right hand screw rule (curl your right hand from direction of $v$ to $B$ and thumb points the direction of force) along $v$ to $B$ (that is $v \times B$) and we see see the thumb point along $-v_{e} Z$ .

Now we also can see that the force is perpendicular to velocity of particle.

Power, if you remember is $F \cdot v$ or $F v \cos \left( \theta \right)$. $\theta = 90^{o}$ so power $= 0$ and work also as power = work/time. So, the particle will perform circular motion.

Here's why 2nd case won't happen. The helical motion is when velocity and electric field makes an acute angle but our $\theta$ here is $90^{o}$. If it's less than $90$, motion goes helical.

Assuming that particle as a charge ,force acting on a charge q is F=q(v x B)

So let say q is along +y axis and E and B (both the fields) along +x axis ,

fine use right hand screw rule (curl your right hand from direction of v to B and thumb points the direction of force) along v to B ( that is v cross B) and we see see the thumb point along -ve Z

Now we also can see that The force is perpendicular to velocity of particle

And power if you remember is F•v or Fv * Cos theta

Theta = 90 so power =0 and work also as power = work/time

So the particle will perform circular motion

Assuming that particle as a charge ,force acting on a charge q is F=q(v x B)

So let say q is along +y axis and E and B (both the fields) along +x axis ,

fine use right hand screw rule (curl your right hand from direction of v to B and thumb points the direction of force) along v to B ( that is v cross B) and we see see the thumb point along -ve Z

Now we also can see that The force is perpendicular to velocity of particle

And power if you remember is F•v or Fv * Cos theta

Theta = 90 so power =0 and work also as power = work/time

So the particle will perform circular motion

And here's why 2nd case won't happen , the helical motion is when velocity and electric field makes an acute angle but our theta here is 90, if it's less than 90 motion goes helical