Calculate the electric potential at point B due to Q. 3.18 times 10^3 V 3.18 tim

Question

Calculate the electric potential at point B due to Q. 3.18 times 10^3 V 3.18 times 10^3 V 3.18 times 10^3 V 3.18 times 10^3 V None What is the magnitude of the electric force between the two charges? 4.50 times 10^6 N 4.50 times 10^9 V 4.50 times 10^3 N 4.50 times 10^8 N None A wire carries a 6.00-A current along the x-axis, and another wire carries a 8.00-A current along the y-axis, as shown in Fig-2. What is the net magnetic field at point P. located at x = 4.00 m. y = 3.00 m? In Fig-3 the battery emf is 50.0 V, the resistance is 250 ohm, and capacitance is 0.50 mu F. The switch S is closed for a long time interval, and zero difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum values of 150 V. What is the inductance? An ac generator with rms voltage of 120-V and frequency of 60.0 Hz is connected in series with a 175.0-ohm resistor, a 15-mu F capacitor and a 50-mH inductor. Calculate the following parameters: (a) angular frequency of the circuit; (b) resonance frequency of the circuit (c) the impedance; and (d) the phase angle.

Explanation / Answer

1. The magnetic field is given by right hand thumb rule .The magnetic field at point P due to the current carrying wire along X-axis is out of the page and the magnetic field at point P due to the current carrying wire along Y-axis is into the page.

Assuming positive direction for direction outwards and negative for direction inwards we have

Btotal = + Bx -By

= (o Ix / 2 y) - (o Iy / 2 x)

= (0 / 2) [(Ix/y) - (Iy/x)]

= (2*10-7Tm/A)[(6A/3m) -(8A/4m)]

= 0

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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