A long straight wire carries a current of 51.7 A. An electron, traveling at 7.43 × 10^7 m/s, is 5.76 cm from the wire.
What is the magnitude of the magnetic force on the electron if the electron velocity is directed (a) toward the wire, (b) parallel to the wire in the direction of the current, and (c) perpendicular to the two directions defined by (a) and (b)?

(a) 2.13*10^{-5} N

(b) 2.13*10^{-5} N

(c) 0 N

Explanation:

the magnitude of the magnetic field generated by the wire is:

if we assume that the current is in the +y direction, B is in the +z direction.

(a) toward the wire, electron is in the -x direction. The angle between B and v is 90°. By using the following formula we obtain:

(b) parallel to the wire, electron is in the +y direction. Again angle between B ans v is 90°.

(c) perpendicular to both previous directions, that is, +z or -z. In this case velocity vector is parallel to the magnetic field vector. Hence:

F3=0N

hope this helps!

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.