A square loop of side 10 cm and resistance 0.7Ω is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, magnitude of induced emf is x√×10−3 V. The value of x is _____
Question
A square loop of side 10 cm and resistance 0.7Ω is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, magnitude of induced emf is x√×10−3 V. The value of x is _____
Solution
The problem involves Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
The formula for Faraday's law is:
emf = -dΦ/dt
where:
- emf is the induced electromotive force,
- dΦ is the change in magnetic flux, and
- dt is the change in time.
Given:
- The side of the square loop, a = 10 cm = 0.1 m,
- The initial magnetic field, B1 = 0.20 T,
- The final magnetic field, B2 = 0 T,
- The change in time, dt = 1 s.
The area of the square loop, A = a^2 = (0.1 m)^2 = 0.01 m^2.
The change in magnetic flux, dΦ = B2A - B1A = (0 T * 0.01 m^2) - (0.20 T * 0.01 m^2) = -0.002 T*m^2.
Substituting these values into Faraday's law gives:
emf = -(-0.002 T*m^2 / 1 s) = 0.002 V = 2 * 10^-3 V.
Therefore, the magnitude of the induced emf is 2√×10−3 V, so the value of x is 2.
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