A 2.0-C Charge moves with a velocity of (2.0i+4.0j+6.0k) m/s and experiences a m

Question

A 2.0-C Charge moves with a velocity of (2.0i+4.0j+6.0k) m/s and experiences a magnetic force of (4.0i-20j+12k)N. It is found that i.B=0. Determine the z component of the magnetic field.

THE ANSWER IS C

Neo-Fle .A É: x = 12 Air A 6-0 sauc F = Q(OXB) vxB OTE, TB2 28, R + (-262 S + C4Bz- 6851 %. 5/2 4. A 2.0-C charge moves with a velocity of (2.0î + 4.0] + 6. Ok) m/s and experiences a magnetic force of (4.0î– 26j + 12k) N. It is found that i. B=0. Determine the component of the magnetic field, A) -3.0T Q- 2.0 C B) +3.0T C) +5.0T V = 2.03+ 4.0S 6.Oks 2010ß HER = (20 145 Meter D) –5.0T E) +6.0T P. 4.0, -205 412k N 21 - 10's tot(48A, -269- ER = Pa- 21 - 1of + GR NESTER 5. A straight 10-cm wire bent at its midpoint so as to form an angle of 90° carries a current of 10 A. It lies in the ry plane in a region where the magnetic field is in the positive z direction and has a constant magnitude of 3,0 ml, What is the magnitude of the magnetic force on this wire? 6 S2 direction a 3.2 mN B) 2.1 mN J: IO A C) 5.3 mN D) 4.2 mN -90° F- BIlsin 10 E) 6.0 mN 1 - . Im Pe 3 mN B= 3.0M TR ke - km AUy =9AV 6. A stream of electrons passes through a velocity filter where the crossed magnetic and electric fields are 0.020 T and 5.00 x 10 V/m, respectively. Find the kinetic energy (in electron volts) of the electrons passing through the filter (1 eV = 1.60 x 10-19 ). Me : 9. JULI leV = 1.6 X 10 J A). 12.2 eV B) 4.85 eV Q:.6 109 C) 5.73 eV B=0.020TU = D) 17.8 ev E) 23,5 ov E = 5.01" VA k: em 2 USE - S vads -- E Sds (7. A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5,0 m on the x-axis. A) 1.6 nt in the negative z direction B), 1.6 nt in the positive z direction F-BIL | -2 C) 2.4 T in the positive z direction D) 2.4 nt in the negative z direction E) None of the above 4r (-0.02 M B-MOI 7- 20 A 1 B = X = 5.0 M Page 2

Explanation / Answer

Force due to magnetic field on a moving particle is given as

F = qv X B

and it is given that x component of B is 0

so lets take B= yj + zk

F = 2 (2.0i+4.0j+6.0k) X (yj +zk) = (4.0i-20j+12k)

comparing both the sides

8z - 12y = 4

-4z = -20

z= +5

hencce Z component of the field is +5T

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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