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The magnetic field of a cylindrical magnet that has a pole-face diameter of 3.3 cm can be varied sinusoidally between 29.6 T and 30.0 T at a frequency of 17 Hz. (The current in a wire wrapped around a permanent magnet is varied to give this variation in the net field.) At a radial distance of 2.5 cm, what is the amplitude of the electric field induced by the variation

Respuesta :

As we know that induced EMF is given by rate of change in magnetic flux

so here we can say

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

here we know that

[tex]\phi = BA[/tex]

so here we can say

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

[tex]EMF = A\times(\Delta B)\times f[/tex]

now by relation of EMF and electric field we can say

[tex]\int E. dl = EMF[/tex]

[tex]E. 2 \pi r = \pi r^2 \times(\Delta B)\times f[/tex]

[tex]E = \frac{r}{2} \times (\Delta B) \times f[/tex]

now plug in all values

r = 2.5 cm

[tex]\Delta B = 30.0 - 29.6 = 0.4 T[/tex]

f = 17 Hz

[tex]E = \frac{0.025}{2} \times 0.4 \times 17 = 0.085 V/m[/tex]

so electric field is given as E = 0.085 N/C

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