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Question: A particle of mass \( 1 \times {10^{ - 26{\text{ }}}}kg \) and charge \( 1.6 \times {10^{ - 19}}C \)...

A particle of mass 1×10−26 kg1 \times {10^{ - 26{\text{ }}}}kg and charge 1.6×10−19C1.6 \times {10^{ - 19}}C travelling with a velocity 1.28×106 ms−11.28 \times {10^6}{\text{ }}m{s^{ - 1}} in the +x+ x direction enters a region in which uniform magnetic field of induction BB are present such that Ex =Ey =0, Ez =−102.4 kVm−1Ex{\text{ }} = Ey{\text{ }} = 0,{\text{ }}Ez{\text{ }} = - 102.4{\text{ }}kV{m^{ - 1}} and Bx = Bz =0, By =8 × 10−2Bx{\text{ }} = {\text{ }}Bz{\text{ }} = 0,{\text{ }}By{\text{ }} = 8{\text{ }} \times {\text{ }}{10^{ - 2}} . The particle enters this region at the origin at time t = 0. Determine the location ( xx , yy and zz coordinates) of the particle at t= 5 × 10−6st = {\text{ }}5{\text{ }} \times {\text{ }}{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 × 10−6 st{\text{ }} = {\text{ }}7.45{\text{ }} \times {\text{ }}{10^{ - 6}}{\text{ }}s ?

Explanation

Solution

In order to solve this question, we are going to consider the Lorentz force for the charge particle, thus finding its magnitude from magnitudes of electric and magnetic fields which come out to be equal and opposite. Given the velocities and time in the two cases, we find the coordinates.
Lorentz force is given by:
F⃗=q(E⃗+v⃗×B⃗)\vec F = q\left( {\vec E + \vec v \times \vec B} \right)
s=vts = vt
Where ss is the distance, vv is the velocity and tt is the time taken.

Complete step by step solution:
Given is the case of a particle travelling with a constant velocity in simultaneous magnetic and electric fields. In such a region, the force acting on the particle is Lorentz force given by:
F⃗=q(E⃗+v⃗×B⃗)\vec F = q\left( {\vec E + \vec v \times \vec B} \right)
Let i^\hat i , j^\hat j and k^\hat k be the unit vectors along the positive directions of xx , yy and zz axes .
QQ is the charge on the particle which equal to 1.6×10−19C1.6 \times {10^{ - 19}}C
vv is the velocity of the charged particle equal to 1.28×106ms−11.28 \times {10^6}m{s^{ - 1}}
The magnitude of the electric field density is −102.4×103Vm−1k^- 102.4 \times {10^3}V{m^{ - 1}}\hat k while the magnitude of the magnetic field is 8×10−2Wbm−2j^8 \times {10^{ - 2}}Wb{m^{ - 2}}\hat j
If Fe{F_e} is the electric force on the charge, then,
{F_e} = qe = \left( {1.6 \times {{10}^{ - 19}}C} \right)\left( { - 102.4 \times {{10}^3}V{m^{ - 1}}\hat k} \right) \\\ \Rightarrow {F_e} = 163.84 \times {10^{16}}N\left( { - \hat k} \right) \\\
If Fm{F_m} is the magnetic force on the charge, then,
Fm=q(v×B){F_m} = q\left( {v \times B} \right)
Putting the values, we get,
{F_m} = \left[ {\left( {1.6 \times {{10}^{ - 19}}C} \right)\left( {1.28 \times {{10}^6}} \right)\left( {8 \times {{10}^{ - 2}}} \right)N} \right]\left( {\hat i \times \hat j} \right) \\\ {F_m} = 163.84 \times {10^{ - 16}}N\hat k \\\
As the electric and magnetic fields are equal and opposite to each other, hence the net force on the charge is zero and there is no deflection.
At time, t=5×10−6st = 5 \times {10^{ - 6}}s
x=5×10−6s(1.28×106)=6.4mx = 5 \times {10^{ - 6}}s\left( {1.28 \times {{10}^6}} \right) = 6.4m
Therefore, coordinates of the particle are (6.4m,0,0)\left( {6.4m,0,0} \right)
When the electric field is switched off, only the magnetic field prevails.
Thus, the particle has uniform circular motion in the x−zx - z plane. In the presence of the magnetic field only, the particle follows a circular path of radius r=mvqBr = \dfrac{{mv}}{{qB}}
r = \dfrac{{\left( {1 \times {{10}^{ - 26}}} \right)\left( {1.28 \times {{10}^6}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {8 \times {{10}^{ - 2}}} \right)}} = 1 \\\ x = 5 \times {10^{ - 6}}s\left( {1.28 \times {{10}^6}} \right) = \\\
The length of arc traced by circle in 7.45 × 10−6 s7.45{\text{ }} \times {\text{ }}{10^{ - 6}}{\text{ s}}
⇒vt=(1.28×106)(7.45 × 10−6 s)=9.536m\Rightarrow vt = \left( {1.28 \times {{10}^6}} \right)\left( {7.45{\text{ }} \times {\text{ }}{{10}^{ - 6}}{\text{ s}}} \right) = 9.536m
Therefore, particle has coordinates
(9.536m,0,2m)\left( {9.536m,0,2m} \right)

Note:
It is important to note that without a magnetic field the path followed by the particle is circular while for under both the fields the path followed is helical. And the coordinates are also found according to the different trajectories in the different states of the electric and the magnetic fields.