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A charged particle carrying chargeq=1 mu C moves in uniform magnetic field with velocity v_1 = 10^6 m/s at angle45^0 with x - axis in x -y plane and experiences a force F_1 = 5 sqrt(2) m N along the negative z-axis . Wen the same particle move with velocity v_2= 10^6 m/s along the z-axis it experiences a force F_2 in y - direction. Find (a) the magnitude and direction of the magnetic field (b) the magnitude of the force F_2. |
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Answer» Solution :`F_2` is in y-direction when velocity is along z-axis Therefore, MAGNETIC field should be along x-axis. So let , `vec B = B_0 hati` a) Given `vec(V_1) = (10^6)/(sqrt(2)) hat i + (10^6)/(sqrt(2)) hat J` and `vec F_1 = - 5 sqrt(2) XX 10^(-3) hatk` From the equation, `vec F = Q(vec v xx vec B)` we have `(-5 sqrt(2) xx 10^(-3)) hat k = (10^(-6))[(10^6/sqrt(2) hat i + (10^6)/(sqrt(2)) hat j) xx (B_0 hat i) ] = - (B_0)/(sqrt(2)) hatk` `therefore (B_0)/(sqrt(2)) = 5sqrt(2) xx 10^(-3) or B_0 = 10^(-2)T` Therefore , the magneticfield is , `vec B = (10^(-2) hati) T` `b) F_2 = B_0 qv_2 " sin " 90^@` as the angle between `vecB` and `vecv` in this case is `90^@` `therefore F_2 = (10^(-2))(10^(-6))(10^6) = 10^(-2)N` |
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