Three points (A, B, and C) make up the vertices of an equilateral triangle with
ID: 1534828 • Letter: T
Question
Three points (A, B, and C) make up the vertices of an equilateral triangle with sides of length equal to 4.0 cm. Two currents are oriented perpendicular to the paper. They intersect the plane of the triangle at the two bottom corners. The currents have equal magnitudes of 0.45 A, but point in opposite directions. (a) Give a symbolic expression for the components of the magnetic field at point C due to the current at A. (b) Give a symbolic expression for the components of the magnetic field at point C due to the current at B. (c) What is the net magnetic field at point C? Give its magnitude and direction with respect to the axes shown. (d) Suppose that a particle with a mass of m = 2.07 times 10^-8 kg and a charge of +1.5 mu C is passing through point C with a velocity of 3.4 times 10^3 m/s at an angle of 30 degree above the +x-direction. (i) What is the net magnetic force felt by the particle? Give both the magnitude and the direction of the force. (ii) What is the instantaneous acceleration of the particle at this point? Again, give both the magnitude and the direction.Explanation / Answer
(a)
magnetic field due to A,
BAx = uo*I*cos30/(2pi*a)) = 4*pi*10^-7*0.45*cos30/(2*pi*0.04) = 1.95*10^-6 T
BAy = -uo*I*sin30/(2pi*a)) = -4*pi*10^-7*0.45*sin30/(2*pi*0.04) = -1.125*10^-6 T
(b)
magnetic field due to B,
BBx = uo*I*cos30/(2pi*a)) = 4*pi*10^-7*0.45*cos30/(2*pi*0.04) = 1.95*10^-6 T
BBAy = +uo*I*sin30/(2pi*a)) = 4*pi*10^-7*0.45*sin30/(2*pi*0.04) = 1.125*10^-6 T
(c)
Bx = BAx + BBx = 3.9*10^-6 T
By = BAy + BBy = 0
B = sqrt(Bx^2+By^2) = 3.9*10^-6 T
(d)
magnetic force Fb = q*(v x B)
v = 3.4*10^3*cos30i + 3.4*10^3*sin30j
v = 2.94*10^3 i + 1.7*10^3 j
B = 3.9*10^-6 i
Fb = 1.5*10^-6*((2.94*10^3 i + 1.7*10^3 j) x 3.9*10^-6 i )
Fb = 1.5*10^-6*1.7*10^3*3.9*10^-6 -k
Fb = -9.945*10^-9 N k
magnitude = 9.945*10^-9
direction along -z axis
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ii)
acceleration a = F/m = 9.945*10^-9/(2.07*10^-8) = 0.48 m/s^2
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