{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW7solutions

# HW7solutions - Physics 2213 HW#7 Solutions Spring...

This preview shows pages 1–2. Sign up to view the full content.

Physics 2213 HW #7 Solutions Spring 2009 F B 30 o 60 o B down up N S #1 [Magnetic Force on an Electron] (a) In order for v B F e   to be up (out of page), v must have a North component. The East- West component won’t affect the force, so it could be anything. There are two possible directions for v : North of West or North of East. (See diagram at right.) 14 vBsin 4 10 N F e 14 19 5 4 10 N 56.4 (1.6 10 C)(5 10 m/s)(0.6T) sin vB F e (b) The greatest magnetic force occurs when the electron is moving perpendicular to the magnetic field. Then F max =evB=(1.6x10 -19 C)(5x10 5 m/s)(0.6T)=4.8x10 -14 N If we want this force to be up, then the electron should be moving due North. (See diagram at right.) #2 [Magnetic Force on a Cable] (a) The force on the cable due to the Earth’s magnetic field is F IL B . The direction will be due South and 30 o below the horizontal (see diagram on right). The magnitude of this force is F B =ILBsin90 o (90 o is the angle between B and L .) =(120A)(30m)(0.5x10 -4 T)(1)=0.18N (b) The weight of the cable is given by F g =mg= Vg=r( r 2 L)g=(8.9x10 3 kg/m 3 )( )(0.0025 2 m 2 )(30m)(9.8N/kg) =51.4N The magnetic force is much smaller than the cable’s weight, so it can be ignored when designing the cables. 27.19. [Fusion Reactor] In part (a), apply conservation of energy to the motion of the two nuclei. In part (b) apply 2 / . q vB mv R (a) Let point 1 be when the two nuclei are far apart and let point 2 be when they are at their closest separation. Apply conservation of energy to the motion of the two nuclei. 1 1 2 2 K U K U . 1 2 0, U K so 1 2 K U . Now U 2 =ke 2 /r , while 2 1 1 2 2 K mv (because there are two deuterium nuclei, each of mass m , moving at speed v ) so 2 2 mv ke r . 19 6 27 15 (1.602 10 C) 8.4 10 m s (3.34 10 kg)(1.0 10 m) k k v e mr (3% of the speed of light!) (b) Recall that a charged particle moving perpendicular to a uniform magnetic field will travel in a circle with 2 / . q vB mv R Then 27 6 19 (3.34 10 kg)(8.4 10 m/s) 0.14 T (1.602 10 C)(1.25 m)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}