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1 answer:
Answer:
68.8 N 13.8°N of W
Explanation:
F₁ is 50 N 30°N of W. The terminal angle is 150°.
F₂ is 25 N 20°S of W. The terminal angle is -160°.
Graphically, you can add the vectors using head-to-tail method. Move F₂ so that the tail of the vector is at the head of F₁. The resultant vector will be from the tail of F₁ to the head of F₂.
Algebraically, find the x and y components of each vector.
F₁ₓ = 50 N cos(150°) = -43.3 N
F₁ᵧ = 50 N sin(150°) = 25 N
F₂ₓ = 25 N cos(-160°) = -23.5 N
F₂ᵧ = 25 N sin(-160°) = -8.6 N
The x and y components of the resultant vector are the sums:
Fₓ = -43.3 N + -23.5 N = -66.8 N
Fᵧ = 25 N + -8.6 N = 16.4 N
The magnitude of the resultant force is:
F = √(Fₓ² + Fᵧ²)
F = √((-66.8 N)² + (16.4 N)²)
F = 68.8 N
The direction of the resultant force is:
θ = tan⁻¹(Fᵧ / Fₓ)
θ = tan⁻¹(16.4 N / -66.8 N)
θ = 166.2°
θ = 13.8°N of W
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Answer:
Explanation:
Given that,
Radius of the circular loop, r = 10 cm = 0.1 m
Current flowing in the loop, I = 3.6 A
Uniform magnetic field, B = 12 T
To find,
The magnetic dipole moment of the loop.
Solution,
Let M is the magnitude of magnetic dipole moment of the loop. We know that the product of current flowing and the area of cross section. Its formula is given by :
A is the area of circular wire
Therefore, the magnetic dipole moment of the loop is . Hence, this is the required solution.