constant, uniform magnetic field (shaded region in the figure). The electron moves in a circular orbit at constant speed in an x-z plane, in the direction shown. It takes the electron 5.2 seconds to complete one orbit. The magnitude of the magnetic field B is Жст) 1.0- 十 -4.0 -3.0 -20 十+ x(cm) 4.0 -10 1.0 3.0 -1.0- 6.9 x 10 12 T. 8.2 x 10 T. 1.1 x 10 12 T. 5.2 x 10 10 T. 4.3 x 10 1 T. Save Al +二

One long wire lies along an x axis and carries a current of 61 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 7.1 m, 0), and carries a current of 42 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0,2.0 m, 0)? Number i Units

Needs Complete typed solution with 100 % accuracy.

One long wire lies along an x axis and carries a current of 33 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.4 m, 0), and carries a current of 64 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 0.80 m, 0)? Number Units the tolerance is +/-2%

A proton moves with a speed of 207258 m/s. It is traveling in the x y plane. A magnetic field of 2.1 mT is pointing in the z-direction. What is the radius (in meters) of the circular path this proton will follow?

a.) The direction of motion of a proton and the B-field are shown by the arrows in the figure. Find both the magnitude and the direction of the magnetic force. b.) Repeat with an electron instead of a proton. Given: Speed of proton/electron: 1.4 x 10^4 m/s and uniform B-field of 2 T.

9. Two long, straight wires are separated by distance, d = 21.0 cm. The wires carry currents of I₁ 2.50 A and I2 = 8.50 A in the same directions, as shown in the figure. Find the magnitude of the net magnetic cm, r2 = field at point (B). Let r₁ = 13.0 7.00 cm, and r3 = 14.0 cm. T r₂ OSS SS 09- A B ssf60 ssf60 ssf60 s60 ssf60 ssf60 ssf60 ssf60 f60 ssf60 ssf60 ssf60 ssf60 ssf60 ssfo d £60 *$$ (9js8 09888 f60 s: 60 ssf60 ssf60 ssf60 ssf60 ssf60 ssf60 f60 ssf f60 ss$60 ssf60 ssf60 ssf60 ssf60 ssf60 ss

What would be the magnitude and direction of force FB on a test positive charge (q = + 3e, me = 29.1 x 10-31 kg) shot in the -y direction with a speed of 7.7 x 105 m/s in a uniform magnetic field B = 2.6 x 10-4 T along +y direction? O FB = 1.97 x 10-¹7 N toward us, away from the plane of the screen O FB = 7.4 x 10-13 N away from us, towards the plane of the screen O FB = 4.92 x 10-11 N away from us, towards the plane of the screen FB = 0.00 N so will maintain its velocity.

At one instant, a proton is at the origin with a velocity (9 × 106, 8 × 106, 4 × 106) m/s. At the observation location (6 × 10-8,3 × 10-8,6 × 10-8) m. A. What is the electric field due to the proton? ( B. What is the magnetic field due to the proton?

A magnet in the form of a cylindrical rod has a length of 5.96 cm and a diameter of 1.28 cm. It has a uniform magnetization of 5.59 × 103 A/m. What is its magnetic dipole moment? Number i Units