Assume that the same amount of current passes through the electromagnets shown. Identify which electromagnet is stronger than th
e electromagnet above.
2 answers:
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Answer:
C_{y} = 4.96 and θ' = 104,5º
Explanation:
To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.
Let's start by decomposing the two vectors.
Vector A
sin θ = / A
cos θ = Aₓ / A
A_{y} = A sin θ
Ax = A cos θ
A_{y} = 4.9 sin 31 = 2.52
Ax = 4.9 cos 31 = 4.20
Vector B
B_{y} = B sin θ
Bx = B cos θ
B_{y} = 6 sin 156 = 2.44
Bx = 6 cos 156 = -5.48
The components of the resulting vector are
X axis
Cx = Ax + B x
Cx = 4.20 -5.48
Cx = -1.28
Axis y
C_{y} = Ay + By
C_{y} = 2.52 + 2.44
C_{y} = 4.96
Let's use the Pythagorean theorem to find modulo
C = √ (Cₙ²x2 + Cy2)
C = Ra (1.28 2 + 4.96 2)
C = 5.12
We use trigonemetry to find the angle
tan θ = C_{y} / Cₓ
θ’ = tan⁻¹ (4.96 / (1.28))
θ’ = 75.5
como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es
θ’ = 180 – tes
θ‘= 180 – 75,5
θ' = 104,5º
Answer:
The magnitude and direction of electric field midway between these two charges is along AB.
Explanation:
Given that,
First charge
second charge
Distance = 20 cm
We need to calculate the electric field
For first charge,
Using formula of electric field
Put the valueinto the formula
Direction of electric field along AB
We need to calculate the electric field
For second charge,
Using formula of electric field
Put the valueinto the formula
Direction of electric field along AO
We need to calculate the net electric field at midpoint
Direction of net electric field along AB
Hence, The magnitude and direction of electric field midway between these two charges is along AB.
