Calculate the magnitude and direction of the magnetic force on an electron moving with a velocity of 5.0 × 104 (?̂) m/s in a uniform magnetic field of 0.67 (?̂) T.

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.23 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Be=

A beam of electrons moving with speed v = 2.6 × 10’m/s enter and then leave the rectangular region of the magnetic field, as shown below. The trajectory of electrons is a quarter of a circle of radius R= 0.020 m. Mass of electron m = 9.11 x 10-31 kg, charge of electron charge q = 1.6 × 10-19 C). The magnitude and direction of the magnetic field are: %3D %3D the region of the magnetic field a. 0.45 T, perpendicular to the page, into the page Ob. 0.74 T, perpendicular to the page, out of the page Oc. 0.45 T, perpendicular to the page, out of the page С. O d. 0.074 T, perpendicular to the page, into the page e. 0.68 T, perpendicular to he page, into the page

A thin 1.65 m long copper rod in a uniform horizontal magnetic field has a mass of 58.0 g. When the rod carries a current of 0.286 A directed perpendicular to the magnetic field, it floats in the magnetic field.   What is the field strength of the magnetic field? Answer in units of T.