27 - MAGNETIC FIELD AND MAGNETIC FORCES 27 27.3. IDENTIFY:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
M AGNETIC F IELD AND M AGNETIC F ORCES 27.3. IDENTIFY: The force F on the particle is in the direction of the deflection of the particle. Apply the right-hand rule to the directions of v and B . See if your thumb is in the direction of F , or opposite to that direction. Use sin F q vB with 90 ° to calculate F . SET UP: The directions of v , B and F are shown in Figure 27.3. EXECUTE: (a) When you apply the right-hand rule to v and B , your thumb points east. F is in this direction, so the charge is positive. (b) 63 sin (8.50 10 C)(4.75 10 m/s)(1.25 T)sin90 0.0505 N F q vB ° EVALUATE: If the particle had negative charge and v and B are unchanged, the particle would be deflected toward the west. Figure 27.3 27.5. IDENTIFY: Apply sin F q vB and solve for v . SET UP: An electron has 19 1.60 10 C q   . EXECUTE: 15 6 19 3 4.60 10 N 9.49 10 m s sin (1.6 10 C)(3.5 10 T)sin60 F v qB  EVALUATE: Only the component sin B of the magnetic field perpendicular to the velocity contributes to the force. 27.9. IDENTIFY: Apply q  F v B  to the force on the proton and to the force on the electron. Solve for the components of B . SET UP: F is perpendicular to both v and B . Since the force on the proton is in the + y -direction, 0 y B and ˆˆ xz BB B = i k . For the proton, ˆ (1.50 km/s) v = i . EXECUTE: (a) For the proton, 33 ˆ ˆ ˆ ˆ (1.50 10 m/s) ( ) (1.50 10 m/s) ( ). x z z q B B q B F = i i k j 16 ˆ (2.25 10 N) , F = j so 16 19 3 2.25 10 N 0.938 T (1.60 10 C)(1.50 10 m/s) z B      . The force on the proton is 27
Background image of page 1

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

View Full DocumentRight Arrow Icon
independent of x B . For the electron, ˆ (4.75 km/s)( ) v = k . 33 ˆ ˆ ˆ ˆ ( )(4.75 10 m/s)( ) ( ) (4.75 10 m/s) x z x q e B B e B   F v B k i k = j  . The magnitude of the force is 3 (4.75 10 m/s) x F e B  . Since 16 8.50 10 N F , 16 19 3 8.50 10 N 1.12 T (1.60 10 C)(4.75 10 m/s) x B   . 1.12 T x B  . The sign of x B is not determined by measuring the magnitude of the force on the electron. 2 2 2 ( 1.12 T) ( 0.938 T) 1.46 T xz B B B   . 0.938 T tan 1.12 T z x B B . 40 ° . B is in the xz -plane and is either at 40 ° from the + x -direction toward the -direction z or 40 ° from the -direction x toward the -direction z . (b) ˆˆ BB B = i k . ˆ (3.2 km/s)( ) v = j . 3 ˆ ˆ ˆ ˆ ˆ ( )(3.2 km/s)( ) ( ) (3.2 10 m/s)( ( ) ) x z x z q e B B e B B   F v B j i k k i . 3 16 16 ˆ ˆ ˆ ˆ (3.2 10 m/s)( [ 1.12 T] [0.938 T] ) (4.80 10 N) (5.73 10 N) e      F = k i i k 2 2 16 7.47 10 N F F F .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

27 - MAGNETIC FIELD AND MAGNETIC FORCES 27 27.3. IDENTIFY:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online