An electron moves through a uniform magnetic field given by B = Bxi+(3.74Bx)j. At a particular instant, the electron has velocity V =(2.88i+4.37j)m/s and the magnetic force acting on it is (2.74x10^-19)k N. Find Bx
B=Bxi+3.74Bxj
V=2.88i+4.37j
F=2.74×10-19k N
F=q (V× B)where B and V are vector
V×B=(3.74×2.88k-4.37k)Bx=6.4012Bxk
F=q×6.4012Bxk=1.6×10^-19×6.4012Bxk
Using given value of F we get
2.74×10^-19k=10.24192×10^-19 Bxk
Bx=2.74/10.24192= 0.26753 T
An electron moves through a uniform magnetic field given by B = Bxi+(3.74Bx)j. At a particular...
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