A particle of mass 1×10−26kg and charge 1.6×10−19C travelling with a velocity of 1.28×106m/s along +ve direction of X-axis enters a region in which a uniform electric field E¯ and B¯ a uniform magnetic field are present such that Ez=−102.4kV/m and B¯y=8×10−2wb/m2 The particle enters this region at origin at time t=0 , then

m=1×10−26kg q=1.6×10−19c v=1.28×106m/s v=1.28×106m/s Ez=−102.4 kvm=−102.4×103vm By¯=8×10−2 wb/m2 Fe=Eq=−102.4×103×1.6×10−19 Fe=−163.84×10−16N Fm=Bqv=8×10−2×1.6×10−19×1.28×106 Fm=16.384×10−15 Fm=163.84×10−16N Fnet=Fm+Fm=−163.84×10−16+163.84×10−16 Fnet=0

A particle of mass 1×10−26kg and charge 1.6×10−19C travelling with a velocity of 1.28×106m/s along +ve direction of X-axis enters a region in which a uniform electric field E¯ and B¯ a uniform magnetic field are present such that Ez=−102.4kV/m and B¯y=8×10−2wb/m2 The particle enters this region at origin at time t=0 , then

m=1×10−26kg q=1.6×10−19c v=1.28×106m/s v=1.28×106m/s Ez=−102.4 kvm=−102.4×103vm By¯=8×10−2 wb/m2 Fe=Eq=−102.4×103×1.6×10−19 Fe=−163.84×10−16N Fm=Bqv=8×10−2×1.6×10−19×1.28×106 Fm=16.384×10−15 Fm=163.84×10−16N Fnet=Fm+Fm=−163.84×10−16+163.84×10−16 Fnet=0

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A particle of mass $1 \times 10^{- 26} kg$ and charge $1.6 \times 10^{- 19} C$ travelling with a velocity of $1.28 \times 10^{6} m/s$ along +ve direction of X-axis enters a region in which a uniform electric field $\bar{E}$ and $\bar{B}$ a uniform magnetic field are present such that $E_{z} = - 102.4 kV/m$ and ${\bar{B}}_{y} = 8 \times 10^{- 2} {wb/m}^{2}$ The particle enters this region at origin at time t=0 , then

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net force on the particle is zero

net force acts in X-Z plane.

net force acts on the particle along the $+ve Z- direction$

net force acts on the particle along the $- ve Z- direction$

answer is C.

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Detailed Solution

$m = 1 \times 10^{- 26} kg q = 1.6 \times 10^{- 19} c v = 1.28 \times 10^{6} m/s v = 1.28 \times 10^{6} m/s E_{z} = - 102.4 k \frac{v}{m} = - 102.4 \times 10^{3} \frac{v}{m} \bar{B_{y}} = 8 \times 10^{- 2} {wb/m}^{2} F_{e} = Eq = - 102.4 \times 10^{3} \times 1.6 \times 10^{- 19} F_{e} = - 163.84 \times 10^{- 16} N F_{m} = Bqv = 8 \times 10^{- 2} \times 1.6 \times 10^{- 19} \times 1.28 \times 10^{6} F_{m} = 16.384 \times 10^{- 15} F_{m} = 163.84 \times 10^{- 16} N F_{net} = F_{m} + F_{m} = - 163.84 \times 10^{- 16} + 163.84 \times 10^{- 16} F_{net} = 0$