CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The force on a charged particle due to electric and magnetic fields is given by F=qE+qv×B. Suppose \vec{E} is along the X-axis and B along the Y-axis. In what direction and with what minim.um speed v should a positively charged particle be sent so that the net force on it is zero?


Solution

Given, F=qE+q(v×B)=0

E=(v×B)

So, the direction of v × B should be opposite to the direction of E. Hence v should be in the +ve yz - plane. 

Again, e = vB sin θ

v=EB sin θ

For v to be minimum, 

θ=900 and so vmin=EB

So, the particle must be projected at minimum speed of EB along '-' ve z -axis 

(θ=900) as shown in the figure so that the force is zero. 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image