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A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm / s . The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be ___________ μV.
Correct answer is '50'. Can you explain this answer?
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A conducting circular loop is placed in a uniform magnetic field of 0....
To find the magnitude of the induced emf in the loop, we can use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux through the loop.
The magnetic flux through a loop is given by the equation:
Φ = B * A
Where Φ is the magnetic flux, B is the magnetic field strength, and A is the area of the loop.
Since the loop is circular, the area can be calculated using the equation:
A = π * r^2
Where A is the area and r is the radius of the loop.
In this case, the radius of the loop is increasing at a rate of 1 mm/s. So, the rate of change of the radius is:
(dr/dt) = 1 mm/s = 0.001 m/s
To find the rate of change of the area, we can differentiate the area equation with respect to time:
(dA/dt) = 2 * π * r * (dr/dt)
Substituting the given values, we get:
(dA/dt) = 2 * π * (0.02 m) * (0.001 m/s) = 0.00004 π m^2/s
Now, we can find the rate of change of magnetic flux:
(dΦ/dt) = B * (dA/dt) = 0.4 T * 0.00004 π m^2/s = 0.000016 π Tm^2/s
Finally, the magnitude of the induced emf in the loop is equal to the rate of change of magnetic flux:
Emf = |(dΦ/dt)| = |0.000016 π Tm^2/s| = 0.000016 π Tm^2/s
Therefore, the magnitude of the induced emf in the loop when the radius is 2 cm is approximately 0.000016 π Tm^2/s.
The magnetic flux through a loop is given by the equation:
Φ = B * A
Where Φ is the magnetic flux, B is the magnetic field strength, and A is the area of the loop.
Since the loop is circular, the area can be calculated using the equation:
A = π * r^2
Where A is the area and r is the radius of the loop.
In this case, the radius of the loop is increasing at a rate of 1 mm/s. So, the rate of change of the radius is:
(dr/dt) = 1 mm/s = 0.001 m/s
To find the rate of change of the area, we can differentiate the area equation with respect to time:
(dA/dt) = 2 * π * r * (dr/dt)
Substituting the given values, we get:
(dA/dt) = 2 * π * (0.02 m) * (0.001 m/s) = 0.00004 π m^2/s
Now, we can find the rate of change of magnetic flux:
(dΦ/dt) = B * (dA/dt) = 0.4 T * 0.00004 π m^2/s = 0.000016 π Tm^2/s
Finally, the magnitude of the induced emf in the loop is equal to the rate of change of magnetic flux:
Emf = |(dΦ/dt)| = |0.000016 π Tm^2/s| = 0.000016 π Tm^2/s
Therefore, the magnitude of the induced emf in the loop when the radius is 2 cm is approximately 0.000016 π Tm^2/s.
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A conducting circular loop is placed in a uniform magnetic field of 0....
The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field.
The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is 2 cm.
The magnetic flux through a circular loop of radius r and area A = πr2 placed in a uniform magnetic field B perpendicular to the plane of the loop is given by:
The induced emf in the loop is given by Faraday's law of electromagnetic induction:
In this case, the radius of the loop is expanding at a constant rate of 10−3 m / s , which means that the rate of change of the area of the loop is:
The magnetic flux through the loop is changing at this rate, and the induced emf in the loop is given by:
Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is 2 cm is 50.24 ≃ 50 μV.
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A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm / s . The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be ___________ μV.Correct answer is '50'. Can you explain this answer?
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