The total force experienced by test charge is Fe+Fm=0.
Fe=−Fm
Fe=−q0v0B0(V×B)
Fe=−q0v0B0[(3i^−j^+2k^)×(i^+2j^−4k^)]
Fe=−qv0B0(14j^+7k^)
The electric field produced by the test charge is ∵E=qfe=−v0B0(14j^+7k^)
In a certain region static electric and magnetic fields exist. The magnetic field is given by B=B0(i^+2j^−4k^). If a test charge moving with a velocity v=v0(3i^−j^+2k^) experiences no force in that region, then the electric field in the region, in SI units, is:
Held on 8 Apr 2017 · Verified 6 Jul 2026.
E=−v0B0(i^+j^+7k^)
E=−υ0B0(3i^−2j^−4k^)
E=v0B0(14j^+7k^)
E=−v0B0(14j^+7k^)
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