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A magnetic field of (4.0 X10^(-3)vec k) ...

A magnetic field of `(4.0 X10^(-3)vec k)` T exerts a force of `(4.0veci +3.0vecj )X 10^(-10)` N on a particle having a charge of `1.0X10^(-9)` C and going in the x-y plane. Find the velocity of the particle.

Text Solution

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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)`
`therefore v_y = 10^2 m//s = 100 m //s`
and `3.0 xx 10^(-10) = -4 v_x xx 10^(-12)`
`therefore v_x = - 75 m//s`
`therefore vec v = -75 hat i + 100 hat j`
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