Define the electric field of a single charge and the magnetic force acting on a moving electric charge using the definitions and formulas in the photos. and thus34leV = 1.602 x 10-19 JU =4TE, r-= (1.440 eV- nm)4TE, r= 1.44 eV1.00 nmSome convenient multiples of the electron-volt arekeV = kilo electron-volt = 10' eVMeV = mega electron-volt = 10 eVFor calculations at the nuclear level, the femtometer is a more convenient unit ofdistance and MeV is a more appropriate energy unit:(1 MeV10 eV= (1.440 eV nm)= 1.440 MeV fmGeV = giga electron-volt = 10° eV10° nmImIt is remarkable (and convenientsame value of 1.440 whether we use typical atomic energies and sizes (eV- nm)or typical nuclear energies and sizes (MeV fm).A magnetic field B can be produced by an electric current i. For example, themagnitude of the magnetic field at the center of a circular current loop of radius ris (see Figure 1.6a)remember) that the quantity e/4xe, has the(In some older works you may find reference to the BeV, for billion electron-volts;this is a source of confusion, for in the United States a billion is 10° while inEurope a billion is 1012 )Ofen we wish to find the potential energy of two basic charges separated bytypical atomic or nuclear dimensions, and we wish to have the result expressedin electron-volts. Here is a convenient way of doing this. First we express thequantity e /4ne, in a more convenient form:(1.11)The SI unit for magnetic field is the tesla (T), which is equivalent to a newton perFIGURE 1.6 (a)A circular current ampere-meter. The constant ao isloop produces a magnetic field B atits center. (b) A current loop withmagnetic moment i in an externalmagnetic field B The field exertsa torque on the loop that will tend to typically produce the current in metallic wires. The direction of B is chosenrotate it so that i lines up with Bet-= (8.988 x 10° N- m/c³)(1.602 x 10-19 C = 2.307 x 10- N- m?Ho = 4x x 10- N-s/C10" nm= (2.307 x 10-28 N - m²)1.602 x 10-19 JleVBe sure to remember that i is in the direction of the conventional (positive) current,opposite to the actual direction of travel of the negatively charged electrons that= 1.440 eV - nmaccording to the right-hand rule: if you hold the wire in the right hand with theWith this useful combination of constants it becomes very easy to calculateelectrostatic potential energies. For two electrons separated by a typical atomicdimension of 1.00 nm, Eq. 1.9 gives The electrostatic force (Coulomb force) exerted by a charged particle q, onanother charge q2 has magnitude1 9 ll92|F =4πεο r(1.8)The direction of F is along the line joining the particles (Figure 1.5). In the SIsystem of units, the constant 1/4TE, has the value1= 8.988 x 10° N - m²/C2The corresponding potential energy isU =1 4192(1.9)4πεο rIn all equations derived from Eq. 1.8 or 1.9 as starting points, the quantity 1/4xE,must appear. In some texts and reference books, you may find electrostaticquantities in which this constant does not appear. In such cases, the centimeter-gram-second (cgs) system has probably been used, in which the constant 1/4TE,is defined to be 1. You should always be very careful in making comparisonsof electrostatic quantities from different references and check that the units areidentical.An electrostatic potential difference AV can be established by a distribution ofcharges. The most common example of a potential difference is that between thetwo terminals of a battery. When a charge q moves through a potential differenceAV, the change in its electrical potential energy AU isAU = qAV(1.10)At the atomic or nuclear level, we usually measure charges in terms of the basiccharge of the electron or proton, whose magnitude is e = 1.602 x 10-19 C. Ifsuch charges are accelerated through a potential difference AV that is a few volts,the resulting loss in potential energy and corresponding gain in kinetic energy willbe of the order of 10-19 to 10-18 J. To avoid working with such small numbers. experience equal and opposite elec-it is common in the realm of atomic or nuclear physics to measure energies inelectron-volts (eV), defined to be the energy of a charge equal in magnitude tothat of the electron that passes through a potential difference of 1 volt:FIGURE 1.5 Two charged particlestrostatic forces along the line joiningtheir centers. If the charges have thesame sign (both positive or both nega-tive), the force is repulsive; if the signsare different, the force is attractive.AU = qAV = (1.602 x 10-19 C)(1 V) = 1.602 × 10-19 J

Question

Define the electric field of a single charge and the magnetic force acting on a moving electric charge using the definitions and formulas in the photos. and thus34leV = 1.602 x 10-19 JU =4TE, r-= (1.440 eV- nm)4TE, r= 1.44 eV1.00 nmSome convenient multiples of the electron-volt arekeV = kilo electron-volt = 10' eVMeV = mega electron-volt = 10 eVFor calculations at the nuclear level, the femtometer is a more convenient unit ofdistance and MeV is a more appropriate energy unit:(1 MeV10 eV= (1.440 eV nm)= 1.440 MeV fmGeV = giga electron-volt = 10° eV10° nmImIt is remarkable (and convenientsame value of 1.440 whether we use typical atomic energies and sizes (eV- nm)or typical nuclear energies and sizes (MeV fm).A magnetic field B can be produced by an electric current i. For example, themagnitude of the magnetic field at the center of a circular current loop of radius ris (see Figure 1.6a)remember) that the quantity e/4xe, has the(In some older works you may find reference to the BeV, for billion electron-volts;this is a source of confusion, for in the United States a billion is 10° while inEurope a billion is 1012 )Ofen we wish to find the potential energy of two basic charges separated bytypical atomic or nuclear dimensions, and we wish to have the result expressedin electron-volts. Here is a convenient way of doing this. First we express thequantity e /4ne, in a more convenient form:(1.11)The SI unit for magnetic field is the tesla (T), which is equivalent to a newton perFIGURE 1.6 (a)A circular current ampere-meter. The constant ao isloop produces a magnetic field B atits center. (b) A current loop withmagnetic moment i in an externalmagnetic field B The field exertsa torque on the loop that will tend to typically produce the current in metallic wires. The direction of B is chosenrotate it so that i lines up with Bet-= (8.988 x 10° N- m/c³)(1.602 x 10-19 C = 2.307 x 10- N- m?Ho = 4x x 10- N-s/C10&#34; nm= (2.307 x 10-28 N - m²)1.602 x 10-19 JleVBe sure to remember that i is in the direction of the conventional (positive) current,opposite to the actual direction of travel of the negatively charged electrons that= 1.440 eV - nmaccording to the right-hand rule: if you hold the wire in the right hand with theWith this useful combination of constants it becomes very easy to calculateelectrostatic potential energies. For two electrons separated by a typical atomicdimension of 1.00 nm, Eq. 1.9 gives The electrostatic force (Coulomb force) exerted by a charged particle q, onanother charge q2 has magnitude1 9 ll92|F =4πεο r(1.8)The direction of F is along the line joining the particles (Figure 1.5). In the SIsystem of units, the constant 1/4TE, has the value1= 8.988 x 10° N - m²/C2The corresponding potential energy isU =1 4192(1.9)4πεο rIn all equations derived from Eq. 1.8 or 1.9 as starting points, the quantity 1/4xE,must appear. In some texts and reference books, you may find electrostaticquantities in which this constant does not appear. In such cases, the centimeter-gram-second (cgs) system has probably been used, in which the constant 1/4TE,is defined to be 1. You should always be very careful in making comparisonsof electrostatic quantities from different references and check that the units areidentical.An electrostatic potential difference AV can be established by a distribution ofcharges. The most common example of a potential difference is that between thetwo terminals of a battery. When a charge q moves through a potential differenceAV, the change in its electrical potential energy AU isAU = qAV(1.10)At the atomic or nuclear level, we usually measure charges in terms of the basiccharge of the electron or proton, whose magnitude is e = 1.602 x 10-19 C. Ifsuch charges are accelerated through a potential difference AV that is a few volts,the resulting loss in potential energy and corresponding gain in kinetic energy willbe of the order of 10-19 to 10-18 J. To avoid working with such small numbers. experience equal and opposite elec-it is common in the realm of atomic or nuclear physics to measure energies inelectron-volts (eV), defined to be the energy of a charge equal in magnitude tothat of the electron that passes through a potential difference of 1 volt:FIGURE 1.5 Two charged particlestrostatic forces along the line joiningtheir centers. If the charges have thesame sign (both positive or both nega-tive), the force is repulsive; if the signsare different, the force is attractive.AU = qAV = (1.602 x 10-19 C)(1 V) = 1.602 × 10-19 J

Accepted Answer

We know,

The electric field is obtained from reformulation of coulomb's law.
Magnetic field exerts…