A magnetic field is aligned perpendicular to the plane of a circular loop of wire with radius 5.0 cm and 20 turns. The magnetic field B changes with time t according to the following function: B 12 exp(-3t) +1.5t +6 T Determine the induced emf in the circular loop of wire as a (i) function of time, and the time at which the induced emf is 0.1 V. (ii) If the magnetic fieldis instead aligned at 30° with respect to the normal of the circular loop of wire, find the induced emf as a function of time.

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aristocles aristocles · Answer 1 · 2019-09-02T03:15:47+02:00

Answer:

[tex]V = 0.157(-36 e^{-3t} + 1.5)[/tex]

[tex]t = 1.24 s[/tex]

Part b)

[tex]EMF = 0.136(-36e^{-3t} + 1.5)[/tex]

Explanation:

magnetic field due to external source is given as

[tex]B = 12 e^{-3t} + 1.5 t + 6[/tex]

area of the loop is given as

[tex]A = \pi r^2[/tex]

[tex]A = \pi(0.05)^2[/tex]

[tex]A = 7.85 \times 10^{-3} m^2[/tex]

Now we have

[tex]\phi = NBAcos0[/tex]

[tex]\phi = (20)(12e^{-3t} + 1.5t + 6)(7.85 \times 10^{-3})[/tex]

[tex]V = \frac{d\phi}{dt}[/tex]

[tex]V = 0.157\frac{d}{dt}(12e^{-3t} + 1.5t + 6)[/tex]

[tex]V = 0.157(-36 e^{-3t} + 1.5)[/tex]

now we need to find the time at which voltage is 0.1 Volts so we have

[tex]0.1 V = 0.157(-36e^{-3t} + 1.5)[/tex]

[tex]t = 1.24 s[/tex]

Part b)

If magnetic field is inclined at an angle of 30 degree with the normal of the loop then

[tex]\phi = NBAcos30[/tex]

now we know that induced EMF is given as

[tex]EMF = \frac{d\phi}{dt}[/tex]

[tex]EMF = NAcos30\frac{dB}{dt}[/tex]

[tex]EMF = (20)(7.85 \times 10^{-3})cos30(\frac{d}{dt}(12e^{-3t} + 1.5t + 6))[/tex]

[tex]EMF = 0.136(-36e^{-3t} + 1.5)[/tex]