Question

# The force on a charged particle due to electric and magnetic fields is given by →F=q→E+q→v×→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=q→E+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.

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