A long wire carrying a 4.5 A current perpendicular to the xy-plane intersects the x-axis at x=−2.0cm. A second, parallel wire carrying a 2.0 A current intersects the x-axis at x=+2.0cm. At what point on the x-axis is the magnetic field zero if the two currents are in the same direction? At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?

mu_0 (4.5 A) / 2 pi ( 0.020m + x) = mu_0(2.0 A ) / 2pi( 0.020m - x) 4.5 ( 0.020m - x) = 2.0 ( 0.020m + x) x = 0.77 cm mu_0 (4.5 A) / 2pi(x+ 0.020m) = mu_0 (2 A) / 2pi(x- 0.020m) x = 5.2 cm

What is the magnetic field at the center of the loop in the figure? (Figure 1)

Field due to the straight part = µ0 i / (2πR) Field due to the circle part = µ0 i / (2R) Add the two to get the answer 4.14*10^-4 Tesla.

You have a 2.1 m long copper wire. You want to make an N-turn current loop that generates a 1.3 mT magnetic field at the center when the current is 2.0 A . You must use the entire wire. What will be the diameter of your coil?

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3×10−11m with speed 2.2×106m/s. According to this model, what is the magnetic field at the center of a hydrogen atom due to the motion of the electron? Hint: Determine the average current of the orbiting electron.

B= 13T

A particle with a charge of q = -5.90 nC is moving in a uniform magnetic field of B⃗ =( -1.20 T ) k^. The magnetic force on the particle is measured to be F⃗ =( −7.60×10−7 N )j^ Can vy, the y component of velocity be determined? Calculate vx. Can vz, the z component of velocity be determined?

YES 7.6*10^-7N / (5.9*10^-9C * 1.20T) = 107 m/s NO

After being accelerated to a speed of 1.82×105 m/s , the particle enters a uniform magnetic field of strength 0.800 T and travels in a circle of radius 33.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q. Find the ratio m/q for this particle.

(33*10^-2cm * 0.800T) / 1.82*10^5 m/s = 1.45 *10^-6 kg/C

The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit in a magnetic field of 8.9×10−2 T at a frequency of 2.5 GHz , and as they do so they emit 2.5 GHz electromagnetic waves. If the maximum diameter of the electron orbit before the electron hits the wall of the tube is 2.1 cm , what is the maximum velocity the electrons may have?

A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes for a medical procedure. Deuterons are nuclei of deuterium, an isotope of hydrogen, consisting of one neutron and one proton, with total mass 3.34×10−27kg. The deuterons exit the cyclotron with a kinetic energy of 6.10 MeV . What is the speed of the deuterons when they exit? If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons' largest orbit, just before they exit? If the beam current is 370 μA how many deuterons strike the target each second?

a) KE = 1/2 mv^2 6.10E6*1.6E-19 = 0.5*3.34E-27*v^2 v= 2.4*10^7m/s b) q v B = mv^2/r r = m v/ q B d = 2 r = 2 m v/q B = 2*3.34e-27*2.4E7/(1.6E-19*1.25)= 0.808 m c) I = q/t = N e/t Not sure if that is A or uA, which is it? if uA N/t = I/e = 370E-6/1.6E-19= 2.31E15

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were accelerated by a voltage of 3.0 kV; the beam was then steered to different points on the screen by coils of wire that produced a magnetic field of up to 0.48 T . What is the speed of electrons in the beam? What acceleration do they experience due to the magnetic field, assuming that it is perpendicular to their path? What is this acceleration in units of g? If the electrons were to complete a full circular orbit, what would be the radius?

(1.6 X 10-19)(3000) = (.48)(9.11 x 10-31)(v2) v = 3.31 X 107 m/s (1.6 X 10-19)(3.31 X 107)(.48) = (9.11 X 10-31)(a) a = 2.78 X 1018 m/s2 which is 2.7 X 1017 g's r = (9.11 X 10-31)(3.31 X 107)/(1.6 X 10-19)(.48) r = 3.92 X 10-4 m which is .390 mm

The aurora is caused when electrons and protons, moving in the earth's magnetic field of ≈5.0×10−5T, collide with molecules of the atmosphere and cause them to glow. What is the frequency of the circular orbit for an electron with speed 1.0×106m/s? What is the frequency of the circular orbit for a proton with speed 5.0×104m/s?

[(1.6*10^-19C)(5*10^-5)] / [2 pi (9.11*10^-31 kg)] =1.39*10^6 Hz [(1.6*10^-19)(5*10^-5)] / [2 pi (1.67*10^-27)] =762 Hz