A proton, at rest in some region of space where there is an electric and magneti
ID: 1404558 • Letter: A
Question
A proton, at rest in some region of space where there is an electric and magnetic field, experiences a net electromagnetic force of 8.0 x 10-13 N pointing in the positive-x direction. If the same proton then moves with a speed of 1.5 x 106 m/s in the positive-y direction, the net electromagnetic force on it decreases in magnitude to 7.5 x 10-13 N, still pointing in the positive-x direction. Find the magnitude and direction of both the electric field and the magnetic field in which this proton is moving.
Explanation / Answer
net force due to electric and magnetic field (also called lorentz force), is given by
F=q*(v cross product B)+q*E
where q=charge of the particle
v=velocity of the particle
B=magnetic field strength
E=electric field strength
when at rest, speed=v=0
then force due to magnetic field B =0
net force=q*E
given that force =8*10^(-13) N
==> 1.6*10^(-19)*E=8*10^(-13)
==> E=5*10^6 N/C
as proton has positive charge and the force is along +ve x direction,
the electric field will also be along +ve x direction.
calculation for magnetic field:
as the proton is moving along y axis, electrostatic field will remain constant i.e. 8*10^(-13) N.
then force due to magnetic field=7.5*10^(-13)-8*10^(-13)=-5*10^(-14) N
-ve sign suggests that force due to magnetic field is along -ve x axis.
now force due to magnetic field 's direction is given by cross product of velocity vector and B vector.
velocity is given to be along +ve y direction.
then B vector has to be along -ve z direction, so that their cross product will be along -ve x axis.
now , magnitude of such force=q*v*B=0.5*10^(-13)
=> 1.6*10^(-19)*1.5*10^(6)*B=0.5*10^(-13)
==> B=0.208 T
hence final answers are:
electric field=5*10^6 N/C. along +ve x axis.
magnetic field=0.208 T, along -ve z axis.
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