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A charged particle carrying charge q = 1μC moves in uniform magnetic field with velocity v_(1) = 10^(6)m/s at angle 45° with x-axis in x-y plane and experiences a force F_(1)= 5sqrt2mN along the negative z-axis. When the same particle moves with velocity, v_(2) = 10^(6) mis along the z-axis it experiences a force F_(2) in y-direction. Find a) magnitude 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, B = `vecB_(0)hati` a) GIVEN `vecV_(1)=(10^(6))/(sqrt2)hati+(10^(6))/(sqrt2)hatj` and `hatvecF_(1)=-5sqrt2xx10^(3)hatk` From the equation, `vecF=q(vecvxxvecB)`, we have `(-5sqrt2xx10^(-3))hatk=(10^(-6))[((10^(6))/(sqrt2)hati+(10^(6))/(sqrt2)hatj)xx(B_(0)hati)]=-(B_(0))/(sqrt2)hatk` `therefore(B_(0))/(sqrt2)=5sqrtxx10^(-3)` or `B_(0)=10^(-2)T` Therefore, the magnetic field is, `vecB=(10^(-2)hati)T` b) `F_(2)=B_(0)qv_(2)sin90^(@)` As the angle between `vecB` and `vecv` in this CASE is 90°. `thereforeF_(2)=(10^(-2))(10^(-6))(10^(6))=10^(-2)N` |
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