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A magnetic field of (4.0 xx 10^(-3) hat k) T exerts a force (4.0 hat i + 3.0 hat j) xx 10^(-10) N on a particle having a charge 10^(-9) C and moving in the x-y plane. Find the velocity of the particle. |
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Answer» Solution :Given Magnetic force `vecF_(m) = (4.0 hati + 3.0 hatj) xx 10^(-10)N` Let veloctiy of the particle in X-y plane be, ` vec v = v_x HAT i + v_y hat J` Then from the relation `vec F_m = q(vec v xx vec B)` we have `(4.0 hati + 3.0 hatj) xx 10^(10) = 10^(-9)[(v_x hat i + v_y hatj) xx (4 xx 10^(-3) hat k)]` `= (4v_y xx 10^(-12) hat i- 4v _x xx 10^(-12) hat j )` COMPARING the coefficient of `hat i` and `hatj` we have, `4 x 10^(-10) = 4 v_y xx 10^(-12)` `thereforev_y = 10^2 m//s = 100 m //s` and `3.0 xx 10^(-10) = -4 v_x xx 10^(-12)` `thereforev_x = - 75 m//s` `therefore vec v = -75 hat i + 100 hat j` |
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