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An electron moves in the X-Y plane with a speed of m/s. Its velocity vector vec v makes an angle 60^(@) with X axis. There is a magetic field of magnitude 10 Weber/m^(2) directed along the Y axis. Calculate the magnetic force vector acting on the electron. (electron charge = – 1.6 xx 10^(-1) coul.) |
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Answer» Solution :`vec V= (V cos 60) cap i + (V sin 60) cap J` = `(10^(6) xx 0.5)cap i +(10^(6)xxsqrt(3)/2) cap j` `vec B=+(10^(-2) cap j``vec F_(mag)=(e) vec V xx vec B=(-1.6 xx 10^(-19))[(5XX10^(5)cap i+5sqrt(3)xx10^(5) cap j)xx(10^(-2))cap j]` `=(-1.6 xx 10^(-19))[(5xx10^(5))(10^(-2))(ixxj)] =(-8xx10^(-16)) cap k` N The force is directed along the negative direction of Z-axis. |
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