A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time) magnetic field of 1.7 tesla. An emf that has a magnitude of 2.6 volt is induced inthis coil because the coil's area A is shrinking. What magnitude of deltaA÷ delta T,which is rate (in m^2/s) at which the area changes?

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A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time) magnetic field of 1.7 tesla. An emf that has a magnitude of 2.6 volt is induced inthis coil because the coil&#039;s area A is shrinking. What magnitude of deltaA÷ delta T,which is rate (in m^2/s) at which the area changes?

user_27qwe · Accepted Answer

To solve this problem, we can use Faraday&#039;s law of electromagnetic induction, which states that the induced electromotive force (emf) in a coil is equal to the rate of change of magnetic flux through the coil.The formula for the induced emf is given by:ε=−dΦdtwhere:- ε is the induced emf- Φ is the magnetic flux through the coil- t is timeIn this case, the emf has a magnitude of 2.6 volts and is induced in the coil due to a shrinking area. The magnetic field is constant and has a magnitude of 1.7 tesla.The magnetic flux Φ through the coil is given by the product of the magnetic field and the area of the coil:Φ=B·Awhere:- B is the magnetic field- A is the area of the coilDifferentiating the equation with respect to time, we get:dΦdt=B·dAdtSubstituting this expression into Faraday&#039;s law, we have:ε=−B·dAdtWe are given that ε = 2.6 volts and B = 1.7 tesla. We need to find the magnitude of dAdt, which represents the rate of change of area with respect to time.Rearranging the equation, we can solve for dAdt:dAdt=−εBSubstituting the given values, we have:dAdt=−2.61.7Evaluating the expression, we can calculate the magnitude of dAdt:dAdt≈−1.529 m^2/s (rounded to three decimal places)Therefore, the magnitude of the rate at which the area changes (dAdt) is approximately 1.529 m2/s.

A planar coil of wire has a single

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