Please help me!
1)1. Two charged point bodies are placed into the opposite on diagonal vertices of a rectangular with sides of 6 and 8 cm in vacuum – Q1=3.2x10-11 C; Q2=-4.267x10-11 C. Calculate the electric field intensity at two other vertices of the rectangular and the point of its diagonals crossing
2)7. A straight conductor of length l=80cm with current I=20A is placed into a uniform magnetic field with magnetic flux density B=1.2T . Calculate the force influencing the conductor of the mutual displacement between current direction and density is α=30o; α=60o; α=90o.
3)8. The coil with L=30mH is supplied with DC voltage of 50V. Calculate resistance of the coil if at the moment when current achieves the value of 10 A the speed of its increasing is 1000A/s
please show at least formulaes
the magnetic potential in the given X v=2xy .find magnetic field (1m,1m)
Sure, I can help you with these questions. Let's go through each of them step by step:
1) To calculate the electric field intensity at two other vertices of the rectangular and the point of its diagonals crossing, we can use the formula for the electric field due to a point charge:
Electric Field (E) = (k * |Q|) / r^2
Where:
- k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2)
- |Q| is the absolute value of the charge
- r is the distance from the charge
First, let's calculate the electric field at the two other vertices of the rectangle.
For one vertex, the distance (r) is the diagonal of the rectangle, which can be found using the Pythagorean theorem:
r = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 cm
Calculating the electric field at that vertex:
E1 = (k * |Q1|) / r^2 = (9 × 10^9 N m^2/C^2) * (3.2 × 10^-11 C) / (10 × 10^-2 m)^2 = 2.88 N/C
Similarly, for the other vertex:
E2 = (k * |Q2|) / r^2 = (9 × 10^9 N m^2/C^2) * (4.267 × 10^-11 C) / (10 × 10^-2 m)^2 = 3.814 N/C
Next, to calculate the electric field at the point where the diagonals cross, we can again use the formula for the electric field. The distance to this point from each charge is the length of the diagonal (same as the r we calculated earlier).
Calculating the electric field at the crossing point:
E_cross = [(k * |Q1|) / r^2] + [(k * |Q2|) / r^2] = (2.88 N/C) + (3.814 N/C) = 6.694 N/C
Therefore, the electric field intensity at the two vertices of the rectangle and the point where the diagonals cross are:
- Vertex 1: 2.88 N/C
- Vertex 2: 3.814 N/C
- Cross point: 6.694 N/C
2) To calculate the force on a conductor due to a magnetic field, we can use the formula:
Force (F) = I * l * B * sin(α)
Where:
- I is the current in the conductor
- l is the length of the conductor
- B is the magnetic flux density
- α is the angle between the direction of the current and the magnetic field
Let's calculate the force for each case:
For α = 30°:
F1 = I * l * B * sin(30°) = (20 A) * (0.8 m) * (1.2 T) * sin(30°) = 9.6 N
For α = 60°:
F2 = I * l * B * sin(60°) = (20 A) * (0.8 m) * (1.2 T) * sin(60°) = 16.64 N
For α = 90°:
F3 = I * l * B * sin(90°) = (20 A) * (0.8 m) * (1.2 T) * sin(90°) = 19.2 N
Therefore, the force on the conductor for α = 30° is 9.6 N, for α = 60° is 16.64 N, and for α = 90° is 19.2 N.
3) To calculate the resistance of the coil, we can use Ohm's law:
Voltage (V) = Current (I) * Resistance (R)
Also, we know that the rate of change of current is equal to the voltage applied divided by the inductance of the coil:
Rate of change of current (dI/dt) = V / L
Where:
- V is the voltage applied
- L is the inductance of the coil
- dI/dt is the rate of change of current
In this case, dI/dt is given as 1000 A/s, the current (I) is given as 10 A, and the voltage (V) is given as 50V.
Rearranging Ohm's law and substituting the variables:
Resistance (R) = V / I = (50 V) / (10 A) = 5 Ω
Therefore, the resistance of the coil is 5 Ω.
I hope this helps! Let me know if you have any further questions.