- #1
DottZakapa
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- Homework Statement
- A 25.0 cm long metal rod lies in the xy-plane and makes an angle of 36.9 with the positive x-axis and an angle of 53.1 with the positive y-axis. the rod is moving in the +x-direction with a speed of 6.80 m/s. the rod is in a uniform magnetic field B= (0.120T)i-(0.220T)j-(0.0900T)k .
What is the magnitude of the emf induced in the rod? indicate in sketch which end of the rod is at higher potential.
- Relevant Equations
- induced emf
Homework Statement: A 25.0 cm long metal rod lies in the xy-plane and makes an angle of 36.9 with the positive x-axis and an angle of 53.1 with the positive y-axis. the rod is moving in the +x-direction with a speed of 6.80 m/s. the rod is in a uniform magnetic field B= (0.120T)i-(0.220T)j-(0.0900T)k .
What is the magnitude of the emf induced in the rod? indicate in sketch which end of the rod is at higher potential.
Homework Equations: induced emf
emf =VxBxL
i have done first
vxB= (6.8 m/s i) x( (0.120T)i-(0.220T)j-(0.0900T)k)= (0.612j -1.496k)
next i split in components the length
0,25(cos 36.9 + sin 36.9)= 0.199 i+0.15j
then
(VxB)x(0.199 i+0.15j)=(0.612j -1.496k)x(0.199 i+0.15j)= (0.224i-0.297j-0.121k)
now getting the magnitude of this i don't get the right result. Can anyone explain to me why?
What is the magnitude of the emf induced in the rod? indicate in sketch which end of the rod is at higher potential.
Homework Equations: induced emf
emf =VxBxL
i have done first
vxB= (6.8 m/s i) x( (0.120T)i-(0.220T)j-(0.0900T)k)= (0.612j -1.496k)
next i split in components the length
0,25(cos 36.9 + sin 36.9)= 0.199 i+0.15j
then
(VxB)x(0.199 i+0.15j)=(0.612j -1.496k)x(0.199 i+0.15j)= (0.224i-0.297j-0.121k)
now getting the magnitude of this i don't get the right result. Can anyone explain to me why?