A square, single-turn coil 0.20 m on a side is placed with its plane perpendicular to a constant magnetic field. And emf of 18mV is induced in the coil winding when the areas of the coil decreases at the rate of 0.10 m^2/s. What is the magnitude of the magnetic field. (Note the change in flux can be due to a change in magnetic field of a change in area. Note that the unit of Tesla can be Volt sec / m^2. )
To find the magnitude of the magnetic field, we can use Faraday's Law of electromagnetic induction:
emf = -dΦ/dt
where emf is the electromotive force induced in the coil, dΦ/dt is the change in magnetic flux through the coil per unit time.
In this case, the change in flux (ΔΦ) is due to the change in the area of the coil (ΔA) and the constant magnetic field (B):
ΔΦ = BΔA
Given that the coil area decreases at a rate of 0.10 m^2/s, we have:
ΔA = -0.10 m^2/s
Substituting these values into the equation above:
emf = -BΔA/dt
18 mV = -B(-0.10 m^2/s) / dt
Now, notice that the unit of magnetic field in the equation is Volt sec / m^2, which is the same as Tesla (T).
Therefore, we can re-write the equation as:
18 mV = B(-0.10 m^2/s) / dt
Solving for B:
B = -(18 mV)(dt) / (-0.10 m^2/s)
To find the value of dt, we require additional information or an assumption about the time interval over which the change in flux occurs. Without this information, it is not possible to determine the exact value of the magnetic field.