A few questions that I‘m having problems with. I need explanations please. For a particle of mass m and charge q moving in a circular path in a magnetif field B, show that its kinetic energy is proportional to r^2, the square of the radius of curvature of its path. A proton moves in a circular path perpendicular to a 1.15 T magnetic field. The radius of its path is 8.4 mm. Calculate the energy of the proton in eV. A magnetic field is perpendicular to the plane of the circut. If the bar has a length of 20 cm, a mass of 1.5g, and is placed in a field of 1.7 T, what constant current flow is needed in order for it to accelerate to 30 m/s in a distance of 1.0 m? A straight 1.00 mm diameter copper wire can just float horizontally in air because of the force of the Earth‘s magnetic field B which is horizontal and of magnitude 5e-5 T. What current does the wire carry?
For a particle of mass m and charge q moving in a circular path in a magnetif field B, let the angular velocity be w. Thus the particle velocity is wr. The kinetic energy of this particle is 0.5*m*v^2 0.5*m*w^2*r^2 A*r^2 The centrapetal force on this proton is: q*v*B m*v^2/r, or v q*B*r/m. Hence the kinetic energy of the proton is: 0.5*m*v^2 0.5*(q*B*r)^2/m 0.5*(1.60e-19*1.15*0.0084)^2/1.67e-27 7.15e-16 (J) The acceleration, a, of this 20cm long bar must satisfy this condition: 2*a*(1.0m) (30 m/s)^2. Thus a 450m/s^2 (Wow!) Let I be the required current. From Newton's low Fma, we have: (1.5e-3)*450 I*(0.2)*1.7 or: I 2.0 (C/s) 2.0 A. Copper density is 8.96 g/cm^3 8.96e3 kg/m^3. Let the required current be I. Also assume the length of this copper wire be L (m). Hence the weight of this piece of copper wire is: pi*(5e-4)^2*L*8.96e3*9.8 (N) 0.0670*L N. The upward force must also be this much for the wire to float in the air: 0.0670*L I*L*B, or: I 0.0670/5e-5 1.38 (kA).
Usually. But I would also note that this is USUALLY a waste of money.
