A 20-cm-diameter circular loop of wire is placed in a 0.57-T magnetic field.
Part A
When the plane of the loop is perpendicular to the field lines, what is the magnetic flux through the loop?
Part B
The plane of the loop is rotated until it makes a 42 ∘ angle with the field lines. What is the angle in the equation ΦB = BAcosθ for this situation?
Part C
What is the magnetic flux through the loop at this angle?
Area loop=PI*radius^2=PI*.1^2 m^2
Area=.01PI m^2
flux=B*area=.0057*PI
"Plane of Loop" I suspect you mean the plane the loop is contained in. Normally, in vectors, we deal with the diretion of the loop to be the NORMAL to that plane. Angle of Plane to flux lines then must be 48deg to flux lines
flux=BAcos48
Part A:
To find the magnetic flux through the loop when the plane is perpendicular to the field lines, we need to use the formula:
ΦB = BA
where ΦB is the magnetic flux, B is the magnetic field, and A is the area of the loop.
Given that the diameter of the loop is 20 cm, we can calculate the radius (r) by dividing the diameter by 2: r = 20 cm / 2 = 10 cm.
Converting the radius to meters, we have r = 10 cm * 0.01 m/cm = 0.1 m.
The area of a circular loop is given by the formula:
A = πr²
So, substituting the value of the radius, we get: A = π(0.1 m)² = π * 0.01 m² = 0.01π m².
Now, we can calculate the magnetic flux using the formula:
ΦB = BA = (0.57 T)(0.01π m²) = 0.0057π T*m².
Therefore, the magnetic flux through the loop when the plane is perpendicular to the field lines is 0.0057π T*m².
Part B:
To find the angle in the equation ΦB = BAcosθ, we need to consider that the loop is rotated until it makes a 42° angle with the field lines.
The angle (θ) in the equation is the angle between the normal to the loop (perpendicular to the plane) and the magnetic field lines.
When the loop is rotated, the angle between the normal to the loop and the field lines is 42°.
So, in this case, θ = 42°.
Part C:
To find the magnetic flux through the loop at this angle, we can use the equation:
ΦB = BAcosθ
We already know the values of B and A from Part A, and we just found the value of θ in Part B.
Substituting the values, we get:
ΦB = (0.57 T)(0.01π m²)cos(42°)
Calculating this expression, we find the magnetic flux through the loop at a 42° angle.
To answer these questions, we need to understand the concept of magnetic flux and its calculation using the equation ΦB = BAcosθ, where ΦB is the magnetic flux, B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field lines and the plane of the loop.
Part A:
When the plane of the loop is perpendicular to the field lines, the angle θ is 0 degrees. The equation becomes ΦB = BAcos0 = BA(1) = BA. Therefore, the magnetic flux through the loop is equal to the product of the magnetic field strength and the area of the loop.
To calculate the area of the loop, we can use the formula for the area of a circle: A = πr^2, where r is the radius of the loop. Given that the diameter of the loop is 20 cm, the radius is half of that, which is 10 cm or 0.1 m.
Substituting the values, we have A = π(0.1)^2 = 0.01π square meters.
Now, we can calculate the magnetic flux through the loop by multiplying the field strength and the loop's area. Given that the magnetic field strength is 0.57 T, the magnetic flux ΦB = (0.57 T)(0.01π square meters).
Part B:
When the plane of the loop is rotated, the angle θ between the magnetic field lines and the plane of the loop changes. In this case, the angle is 42 degrees, as mentioned in the question. We need to determine the angle θ in the equation ΦB = BAcosθ for this situation.
Part C:
To answer Part C, we need to calculate the magnetic flux through the loop at this new angle. Using the same equation ΦB = BAcosθ, we can substitute the values we already know: B = 0.57 T, A = 0.01π square meters, and θ = 42 degrees.