It will produce an electric field and a magnetic, both uniform and crossed at 90

Question

It will produce an electric field and a magnetic, both uniform and crossed at 90 degrees to each other to use as a gear selector of charged ions. The electric field will get a capacitor, C : parallel plate , while the magnetic field will be obtained with two electromagnets , LI and L2 , one against another, whose iron nuclei far so little that the magnetic field between the two cores is practically the same as the interior of the solenoid Feeding circuit's capacitor and electromagnet are independent and are shown schematically in Fig. Two. The velocity of the ions to be selected is 6.0 times 10 ^ 6 m / s, so it takes the fields are chosen values? E = 3.0 times 10 ^ 4 V / m and B = 0.0050 T (thus, v = E / B = 6.0 times 10 ^ 6 m/ s) If the capacitor has plates of 4.0cm ^ 2 and a plate spacing of 0.50cm: i- What voltage must be applied between the plates? That current must pass through the resistor in parallel with the capacitor voltage for attaining this ? If this stream has to achieve by adjusting the variable resistor R, to the resistor value should be adjusted ? Neglect the resistance of the source, whose electromotive force is 200v The velocity of the ions to be selected is 6.0 times 10 ^ 6 m / s, so it takes the fields are chosen values? E = 3.0 times 10 ^ 4 V / m and B = 0.0050 T (thus, v = E / B = 6.0 times 10 ^ 6 m/ s) If the capacitor has plates of 4.0cm ^ 2 and a plate spacing of 0.50cm: i- What voltage must be applied between the plates? That current must pass through the resistor in parallel with the capacitor voltage for attaining this ? If this stream has to achieve by adjusting the variable resistor R, to the resistor value should be adjusted ? Neglect the resistance of the source, whose electromotive force is 200v

Explanation / Answer

The voltage on the capacitor is coming from the definitions of electric field E

E =U/d so that

U =E*d =3*10^4*0.5*10^-2 =150 V

The current on the resitor in parallel (the voltage on resitor is the same) is just

I =U/R =150/300 =0.5 A

Assumin the capacitor is fully charged this curent is the same also through variable resitor. The volatge on the variable resitors i

V(R) = Vcc-U =200-150 =50 V

Rvar =V(R)/I =50/0.5 =100 Ohm

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.