Four long, parallel conductors carry equal currents of I = 5.00 A. The direction of the current is into the page at points A and
B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square with an edge length 0.200 m.
1 answer:
Answer:
Explanation:
The magnetic field due to a straight wire carrying current is given by
B = μo• I / 2πr
Where,
μo = permeability of free space
μo = 4π×10^-7 Tm/A
I is the current in wire
I = 5A
r Is the distance of point from the wire.
The distance between the parallel conductors and the point is r/2
R = 0.283/2 = 0.1415m
Check attachment on how I used Pythagoras theorem to find the diagonal of the square..
Hence, the magnetic field at point P is
B = μo•I / 2πR
B = 4π × 10^-7 × 5 / 2π × 0.1415
B = 7.07 × 10^-6 T.
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Answer:
The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
Explanation:
Here is the complete question
Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.
Solution
The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by
F = kq₁q₂/r₁²
q₁q₂ = Fr₁²/k
q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²
q₁q₂ = 6 × 10⁻¹⁶ C² (1)
When the charges are brought together, the charge is now q = (q₁ + q₂)/2
The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then
F₂ = kq²/r₂²
q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²
q² = 6.25 × 10⁻¹⁶ C²
q = √(6.25 × 10⁻¹⁶) C
q = 2.5 × 10⁻⁸ C
(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C
(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C
q₁ + q₂ = 5 × 10⁻⁸ C (2)
q₁ = 5 × 10⁻⁸ C - q₂ (3)
Substituting equation (3) into (1), we have
(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²
Expanding the bracket, we have
(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²
So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0
Using the quadratic formula to find q₂
q₁ = 5 × 10⁻⁸ C - q₂
q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C
q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C
So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C
Potential energy is defined by formula
here
m = mass
g = acceleration due to gravity
h = height
Now here two different stones are located at same height
while mass of stone A is twice that of stone B
so here we can say potential energy of A is
Similarly potential energy of B is
now if we take the ratio of two energy
so we can say potential energy of stone A is two times the potential energy of B