In a certain region static electric and magentic fields exist. The magnetic field is given by `vecB=B_(0)(hati+2hatj-4hatk)` . If a test charge moving with a velocity `vecu=-u_(0)B_(0)(3hati-2hatj-4hatk)` experience no force in that region, then the electric field in the region, in SI units, is-

Options

A. `vecE=-v_(0)B_(0)(14 hatj+7hatk)`
B. `vecE=v_(0)B_(0)(14hatj+7hatk)`
C. `vecE=-v_(0)B_(0)(hati+hatj+7hatk)`
D. `vecE=-v_(0)B_(0)(3i-2hatj-4hatk)`

Answer

A

Explanation

Here `vecB=B_(0)(hati+2hatj-4hatk), hatv=v_(0)(3hati-hatj+2hatk), vecF=vecF_(e)+vecF_(m):.vecF=0` or `vecF_(3)=-vecF_(m)`
`vecF_(e)=-q(vecv xx vecB)=-qv_(0)B_(0)[(3hati-hatj+2hatk)xx(hati+2hatj-4hatk)]=-qv_(0)B_(0)(14hatj+7hatk)`
The electric field produced by the charge q, will be,
`vecE=(vecF_(e))/q=(-qv_(0)B_(0)(14hatj+7hatk))/q` or `hatE=-v_(0)B_(0)(14hatj+7hatk)`