Respuesta :
Answer:
(a) [tex]F=4.3697\times 10^{-3}\,N[/tex]
(b) [tex]q= 1.6583\times 10^{-4} \,C[/tex]
Explanation:
Given:
- mass of positive charge, [tex]m=7.2\times 10^{-8}\,kg[/tex]
- speed of charge, [tex]v=85\,m.s^{-1}[/tex]
- magnetic field, [tex]B=0.31 \,T[/tex]
- time, [tex]t=2.2\times 10^{-3} \,s[/tex]
According to the given condition the positive charge enters due east and leaves the magnetic field heading due south completing one-quarter of a circle during the course of time during which it travels perpendicular to the magnetic field.
Now, by the Fleming's left hand rule we obtain that the direction of the field is outward of the plane in which the charge has traced the locus.
Firstly we find the angular velocity:
[tex]\omega= \frac{\theta^c}{t} \,rad.s^{-1}[/tex]
[tex]\omega = \frac{\frac{\pi}{2} }{2.2\times 10^{-3}}[/tex]
[tex]\omega=713.998 \,rad.s^{-1}[/tex]
We know,
[tex]F=m.r.\omega^2[/tex].........................(1)
also
[tex]F=q.v.B[/tex].......................................(2)
from eq.(1) & (2)
(b)
[tex]m.r.\omega^2=q.v.B[/tex]
â”[tex]\omega=\frac{v}{r}[/tex]
[tex]m.\omega=q.B[/tex]
Putting the respective values in the above expression:
[tex]7.2\times 10^{-8}\times 713.998 =q\times 0.31[/tex]
[tex]q= 1.6583\times 10^{-4} \,C[/tex]
(a)
Putting the respective values in eq. (2)
[tex]F=1.6583\times 10^{-4}\times 85\times 0.31[/tex]
[tex]F=4.3697\times 10^{-3}\,N[/tex]
