A beam of particles, each with a charge of 8.0 x 10 19 C, moves at [ (-3.43) +(0

Question

A beam of particles, each with a charge of 8.0 x 10 19 C, moves at [ (-3.43) +(0.78j(9.68)k 1x 105 m/s through a uniform magnetic field of I (-2.51)i (7.58)j (-3.69)k ] T. Let F represent the force vector on each particle. Small number entry: use 1.24e-14 N to enter 1.24x10-14 N NOTES | | ?MAGES : Discuss UITS- STATS-HELP Part Description Answer Chk History A:-610.OE 14 N C: -6.100 pN What is the i component of F? (include units with answer) # tries: 2-Show Details Hints: 0.0 295.63E 14 N C: -2.956 Hints: 0. pN What is the j component of F? (include units with answer) # tries: 1 Show Details -192.33E- what is the k component of F? (include units with answer) | ?N # tries: 1 | ShowDetals C: -1.923 pN Hints:-0.0 What is the magnitude of F? (include units with answer) A0512 C: 7.046 pN Hints 1.0 # tries: 5 | Show Detals A uniform electric field of [ (-9.26)i + (-9.36)j +(-9.06)k ] x 105 N/C is added to the system. What is the new magnitude of F? (include units with answer) 10.56 pts.MA| # tries: 2 , show Details , 106% Format Check 2% try penalty Hints: 1.0

Explanation / Answer

Electrostatic force Fe = qE = 8e-19*-9.26e5 i + 8e-19*-9.36e5j + 8e-19*-9.06e5 k

= -7.40800e-13 i   -7.488e-13 j -7.248e-13 k

Net force = [-6.1e-12 -7.40800e-13 ] i   + [-7.488e-13- 2.956e-12] j + [ -7.248e-13 - 1.923e-12] k

= -6.8408e-12 i -3.7048e-12 j -2.6478e-12 k

magnitude = sqrt(6.8408^2+3.7048^2+2.6478^2) *10^-12 N

= 8.218 *10^-12 N

= 8.218  pN answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.