Answer:
The net magnetic field ta the center of square is
.
Explanation:
Current, I = 12 A , side ,a = 10 cm = 0.1 m
Let the magnetic field due to the one side is B.
The magnetic field is given by
Net magnetic field at the center of the square is
B' = 4 B
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃
We divide by (k * q3) on both sides of the equation
q₁ = + 1.25 nC
The electric force between two charged particles can be increased by decreasing the distance between the two particles.
<h3>How to increase electric force between two charged particles.</h3>
The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while
increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.
Read more on Electric Force:
brainly.com/question/17692887
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Answer:
Its inductance L = 166 mH
Explanation:
Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR
R = V/I = 5.62/0.698 = 8.052 Ω
Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from
V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'
Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω
WE now find the reactance X of the coil from
Z² = X² + R²
X = √(Z² - R²)
= √(97.5² - 8.05²)
= √(9506.25 - 64.8025)
= √9441.4475
= 97.17 Ω
Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.
L = X/2πf
= 97.17/2π(93.1 Hz)
= 97.17 Ω/584.965 rad/s
= 0.166 H
= 166 mH
Its inductance L = 166 mH
