need a help 7-12 7. If E1 4.0 V, E 2 12.0 V, R 4 2, R2 12 2, C 3 AuF, Q 18 HC, a
ID: 1267288 • Letter: N
Question
need a help 7-12
7. If E1 4.0 V, E 2 12.0 V, R 4 2, R2 12 2, C 3 AuF, Q 18 HC, and I 2.5 A, what is the potential difference Va Vo? a. 30 V b. 30 V C. 5.0 V 5.0 V 8. A positive point charge enters a uniform magnetic field directed to the right side of the screen, as shown. If the charge is moving to the top of the screen, what is the direction of the magnetic force on the charge? a. into the screen out of the screen b. The force is zero c. to the right side of the screen d. to the left side of the screen 9. A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits? (Hint: 1 eV refers to the kinetic energy of the electrons) a. 2.8 b. 1.7 c. 1.3 d. 4.0 e. 1.0 10. A long straight wire carries a current of 12 A. A uniform magnetic field makes an angle of 53 with the wire, and has a magnitude of 1.6 T. What is the force on a 2.5 m length of wire? d. 38 N e. 24 N C. 29 N b. 64 N a. 15 N 11. A square loop of wire, 40 cm x 40 cm, is hinged on the z-axis as shown. The current in the loop is i 20 A. A uniform magnetic field of 0.8 T is directed in the +x direction. What angle between the x axis and the plane of the loop will produce a torque of 2 N.m on the loop by the magnetic field? a. 18.2 degrees b. 38.0 degrees c. 71.8 degrees d. 38.6 degrees e. 51.4 degrees 12. The algebraic sum of the changes of potential around any closed circuit loop is a. zero b. maximum. c. zero only if there are no sources of emf in the loop. d. equal to the sum of the currents in the branches of the loop.Explanation / Answer
7)
Va -4 + 4*2.5 -12 + 18/3 + 12*2.5 = Vb
Va - Vb = 4 -10 + 12 -6 -12*2.5= - 30 V
8) v into B
into the page
so a)
9)
q v B = m v^2/r
q B = m v/r
r = mv/q B
KE = 1/2 mv^2
v = sqrt( 2 KE/m)
so
r2/r1 = v2/v1 = sqrt(KE2/KE1) = sqrt(5/3) = 1.3 so c)
10)
F = I L B sin theta = 12*1.6*2.5*sin(53 degrees)= 38 N so d)
11)
torque = N I A B cos theta
2 = 1*20*.4^2*0.8*cos(theta)
theta = arcos(2/(1*20*.4^2*0.8))= 38.6 so d)
12. zero so a)
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