A proton traveling at 4.40 suddenly enters a uniform magnetic field of 0.740 , t

Question

A proton traveling at 4.40 suddenly enters a uniform magnetic field of 0.740 , traveling at an angle of 55.0 with the field lines (see the figure (Figure 1) ). a) Find the direction of the force this magnetic field exerts on the proton. b) Find the magnitude of the force this magnetic field exerts on the proton. c)If you can vary the direction of the proton's velocity, find the magnitude of the maximum and minimum forces you could achieve. d)How the velocity should be oriented to achieve the forces in part (C).

Explanation / Answer

s = (1/2)*a*t^2 a = 6.6 m/s^2 F = ma = 4.5*6.6 = 29.7 N The law of gravitation states the the force between any two masses m1 and m2, seperated by distance r, is F = G*m1*m2 / r^2 , where G = 6.67*10^(-11) N*(m^2) / (Kg^2) is the gravitational constant. Assuming that m1 lies at the top corner of the triangle and that the side connecting m2 and m3 is parallel to the x-axis, there are two forces acting on m1 The force between m1 and m2 : F(1,2) = G*m1*m2 / r^2 = 6.67*10^(-11)*40.00*45.00 / 6.0^2 = 3.3*10^(-9) N at a 60 degrees angle The force between m1 and m3 : F(1,3) = G*m1*m3 / r^2 = 6.67*10^(-11)*40.00*85.00 / 6.0^2 = 6.3*10^(-9) N at a 120 degrees angle. Analyzing the two force vectors into two components, one parallel to the x-axis and the other parallel to the y axis we get Fy(1,2) = F(1,2)sin(60) = 3.3*10^(-9)*0.87 = 2.87*10^(-9) N Fx(1,2) = F(1,2)cos(60) = 3.3*10^(-9)*0.50 = 1.65*10^(-9) N Fy(1,3) = F(1,3)sin(120) = 6.3*10^(-9)*0.87 = 5.48*10^(-9) N Fx(1,3) = F(1,3)cos(120) = 6.3*10^(-9)*(-0.5) = -3.15*10^(-9) N So, the net force on the y-axis is Fy = Fy(1,2)+Fy(1,3) = 8.35*10^(-9) N and on the x-axis Fx = Fx(1,2)+Fx(1,3) = -1.50*10^(-9) N The magnitute of the net force is F = sqrt(Fx^2 + Fy^2) = 8.48*10^(-9) N The angle ? is, tan(?) = Fy/Fx = -5.57 ----> ? = 100 degrees.

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