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A magnetic field of ( 4.0 xx 10^(-3)hatk) T exerts a force ( 4.0hati + 3.0hatj ) 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 :Magnetic force `vecF_(m)=(4.0hati+3.0hatj)xx10^(-10)N` Let velocity of the particle in X-y plane be,`vecv=v_(x)hati+v_(y)hatj` Then from the relation `vecF_(m)=1(vecvxxvecB)` We have `(4.0hati+3.0hatj)xx10^(-10)=10^(-9)(v_(x)hati+v_(y)hatj)xx(4xx10^(-3)hatk)]=(4v_(y)xx10^(-12)hati-4v_(x)xx10^(-12)hatj)` . comparing the COEFFICIENT of `hati` and `hatj` we have, `4xx10^(-10)=4v_(y)xx10^(12)` `thereforev_(y)=10^(2)m//s=100m//s` and `3.0xx10^(-10)=-4v_(x)xx10^(-12)` `thereforev_(x)=-75m//sthereforevecv=-75hati+100hatj` |
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