Explanation:
It is given that, the field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs
Change in magnetic field, [tex]dB=2.5\ T[/tex]
Change in time, [tex]dt=200\ \mu s=200\times 10^{-6}\ s=2\times 10^{-4}\ s[/tex]
Diameter, d = 2.3 cm
Radius, r = 0.0115 m
Emf is induced in the ring as the field changes. It is given by :
[tex]E=\dfrac{d\phi}{dt}[/tex]
[tex]E=\dfrac{d(B.A\ cos(0))}{dt}[/tex]
[tex]E=A\dfrac{d(B)}{dt}[/tex]
[tex]E=\pi (0.0115)^2\dfrac{2.5}{2\times 10^{-4}}[/tex]
E = 5.19 volts
So, the emf induced in the ring is 5.19 volts. Hence, this is the required solution.
The emf is induced in the ring at the given field strength is 5.2 V.
The emf induced in the ring is determined by applying Faraday's law of electromagnetic induction as follows;
emf = dФ/dt
where;
emf = BA/t
where;
A = πd²/4
A = π x (0.023²)/4
A = 4.16 x 10⁻⁴ m²
emf = (2.5 x 4.16 x 10⁻⁴)/(200 x 10⁻⁶)
emf = 5.2 V
Thus, the emf is induced in the ring at the given field strength is 5.2 V.
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