A charged particle q is moving with a velocity v1=2i^m/s at a point in a magnetic field B and experiences a force F1=q(k^−2j^)N. If the same charge moves with velocity v2=2j^m/s from the same point in that magnetic field and experiences a force F2=q(2i^+k^)N, the magnetic induction at that point will be :
A
i^+21j^−21k^
B
−21i^+21j^+k^
C
21i^+21j^+k^
D
−21i^+j^+21k^
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Updated on : 2022-09-05
Solution
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Correct option is B)
Let the magnetic induction at required point be Bˉ=Bxi^+Byj^+Bzk^
Magnetic force (Fˉ) on particle of charge (q), moving with velocity Vˉ in the magnetic field region of intensity (Bˉ) is given as
Fˉ=q(Vˉ×Bˉ)
Case 1: F1ˉ=q(V1ˉ×Bˉ)
q(k^−2j^)=q(2i^×(Bxi^+Byj^+Bzk^))
k^−2j^=2Byk^−2Bzj^
∴2By=1,By=1/2.....(i)
−2=−2Bz,∴Bz=1.....(ii)
Case 2: F2ˉ=q(V2ˉ×Bˉ)
q(2i^+k^)=q(2j^×(Bxi^+Byj^+Bzk^))
2i^+k^=−2Bxk^+2Bzi^
∴−2Bx=1,∴+Bx=1/2......(iii)
Thus from (i),(ii)&(iii)
Bˉ=−21i^+21j^+k^
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