a)when a particle of positive charge q and velocity v is in a magnetic field B,

Question

a)when a particle of positive charge q and velocity v is in a magnetic field B, what is th eforce on the particle?
b) Using the result of (a) prove that a wire length element dl with cross section area A, and current I normal to A placed in B experiences a force F = i dl * B.
c) Using Ampere's Law calculate the magnetic field producesd at a perpendicular distance r from and infinitely long wire carrying current i.
d) The Ampere A is defined to be the value of the current in each of two parallel straight wires placed 1m apart which produce a force equal to 2x10^-7 N per meter of the wire. Using the results of b and c, show that the magnetic permeability constant has value of mu = 4pi *10^-7 N/A^2

Explanation / Answer

a) Force on a particle due to a magnetic field: F = qvBsinx where q is the charge where v is the velocity of the particle where B is the magnitude of the strength of the magnetic field where x is the angle the velocity makes with the direction of the magnetic field b) F = qvBsin(x) x is 90 degrees because it is placed normal to A (normal means perpendicular) F = qvBsin(90) sin(90) = 1 F = qvB v is defined as dl/t F = q(dl/t)(B) q/t is defined as current F = i(dl)(B) c) B = (mu)(I)/(2pir) B = mu(i)/(2pir) d) F = i(dl)(B) F = i(dl)(mu)(i)/(2pir) F/dl = i(mu)(i)/(2pir) 2x10^-7 = (i^2)(mu)/(2pi*1) 4pix10^-7 = i^2(mu) mu = 4pix10^-7 N/A^2 BOL

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