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Alexus [3.1K]
3 years ago
7

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has

a magnitude of 2.1 T and is inclined at an angle of 65° with respect to the normal to the plane of the loop.
(a) If the magnetic field decreases to zero in a time of 0.45 s, what is the magnitude of the average emf induced in the loop?
(b) If the magnetic field remains constant at its initial value of 2.1 T, what is the magnitude of the rate
ΔA/Δt
at which the area should change so that the average emf has the same magnitude as in part (a)?
Physics
1 answer:
Pani-rosa [81]3 years ago
5 0

Answer:

Part a)

EMF = 0.38 V

Part b)

\frac{dA}{dt} = 0.43 m^2/s

Explanation:

Part a)

Initial value of magnetic flux is given as

\phi_1 = BAcos\theta

\phi_1 = (2.1)(0.35 \times 0.55) cos65

so we have

\phi_1 = 0.17 Wb

Final flux through the loop is given as

\phi_2 = 0

now EMF is given as

EMF = \frac{\phi_1 - \phi_2}{\Delta t}

EMF = \frac{0.17 - 0}{0.45}

EMF = 0.38 V

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

E = Bcos65 \times \frac{dA}{dt}

0.38 = 2.1 cos65(\frac{dA}{dt})

\frac{dA}{dt} = 0.43 m^2/s

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