An infinitely long wire carries current I = 5 A. At some instant, a charge q = -
ID: 1490100 • Letter: A
Question
An infinitely long wire carries current I = 5 A. At some instant, a charge q = -2 Times 10^-4 C has instantaneous velocity v = 200 m/s parallel to the current d = 20cm away from it. What is the magnetic force (both magnitude and direction) acting on the charge at this moment How far from the charge should a second infinitely long wire carrying a current I = 1.5 A be, so that the charge will maintain its velocity constant? ( Clue: Constant velocity means that acceleration is zero and so is the net force in the charge)Explanation / Answer
a) We know that, Mgnetic force,F is given by:-
F= q(v*B*sin(theta))---1 where,q=charge,v=velocity of charge,B=Magnetic field acting on it
and theta= angle between magnetic field and velocity
In this case, theta= 90 degree
Let's find B:-
B= meu(o)*i/(2*pi*r) where, i=current, d= given seperation,meu(o)=4Pi*10^(-7)
So, B= 2*10^(-7) *5/(0.2) = 0.5*10^(-5) T
Putting this value in eq 1,we get,
F= -2*10^(-4) *200*0.5*10^(-5) = -2*10^-7 N
b) The value of F which we got in the previous part is the value of force exerted
by the second wire but in opposite direction.
Hense B will be same as in previous case.
So, 0.5*10^(-5)= meu(o)*i/(2*pi*r) = 2*1.5*10^-7/r
=> r= 6*10^(-2) meter = 6 cm.