# A particle of mass 1 x 10 -26 kg and charge + 1.6 x 10 -19 coulomb travelling with a velocity 1.28 x 10 6 m/s in the +X direction enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present such that E x = E y = 0, E z = - 102.4 kV / m and B x = B z = 0, B y = 8 x 10 -2 water/ m 2 . The particle enters this region at the origin at time t = 0. Determine that location (x, y and x coordinates ) of the particle at t = 5 x 10 -6 s. If the electric field is switched off at this instant (with the magnetic field still present ), what will be the position of the particle at t = 7. 45 x 10 -6 s ?

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A particle of mass 1 x 10^{-26} kg and charge + 1.6 x 10^{-19} coulomb travelling with a velocity 1.28 x 10^{6} m/s in the +X direction enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present such that E_{x } = E_{y} = 0, E_{z} = - 102.4 kV / m and B_{x} = B_{z} = 0, B_{y} = 8 x 10^{-2} water/ m^{2}. The particle enters this region at the origin at time t = 0. Determine that location (x, y and x coordinates ) of the particle at t = 5 x 10^{-6} s. If the electric field is switched off at this instant (with the magnetic field still present ), what will be the position of the particle at t = 7. 45 x 10^{-6} s ?

^{-26}kg and charge + 1.6 x 10

^{-19}coulomb travelling with a velocity 1.28 x 10

^{6}m/s in the +X direction enters a region in which a uniform electric field E and a uniform magnetic field of induction B are present such that E

_{x }= E

_{y}= 0, E

_{z}= - 102.4 kV / m and B

_{x} = B

_{z}= 0, B

_{y}= 8 x 10

^{-2}water/ m

^{2}. The particle enters this region at the origin at time t = 0. Determine that location (x, y and x coordinates ) of the particle at t = 5 x 10

^{-6}s. If the electric field is switched off at this instant (with the magnetic field still present ), what will be the position of the particle at t = 7. 45 x 10

^{-6}s ?

## 1 Answers

^{-26}kg, q = + 1.6 x 10

^{-19}C

^{6}m/s) , = - 102.4 x 10

^{3}m/s

^{-2}

_{1}= qE in the – z direction

^{-19}x 102.4 x 10

^{3}= 163. 84 x 10

^{-16}N

_{2}= qvB in the + z direction

^{-19}x 1.28 x 10

^{6}x 8 x 10

^{-2}= 163.84 x 10

^{-16}N

^{6}x 5 x 10

^{-6}= 6.4 m

^{-26 }x 1.28 x 10

^{6}/ 1.6 x 10

^{-19}x 8 x 10

^{-2}= 1m

^{6}= 44.9 x 10

^{-6}sec ….(i)

^{-6}s – 5.0 x 10

^{-6}s ) = 2.45 x 10

^{-6}sec …… (ii)

^{-6}sec will be (6.4 m, 0,2 m)

Navjot Kalra

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