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3 years ago

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A long solenoid has a diameter of 11.1 cm. When a current i exists in its windings, a uniform magnetic field of magnitude B = 42

.9 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 6.65 mT/s. Calculate the magnitude of the induced electric field (a) 4.34 cm and (b) 7.46 cm from the axis of the solenoid.

1 answer:

Pepsi [2]3 years ago

6 0

Answer:

$E(1) = 1.44\times 10^{-4} v/m$

$E(2) = 1.34\times 10^{-4} v/m$

Explanation:

Given data:

diameter of solenoid is 11.1 cm

B = 42.9 mT

dR = 6.65 mT/s

d(1) = 4.34 cm

d(2) = 7.46 cm

we know that

electric field due to solenoid is given as

E = 1/2 dB/dt (r)

$E(1)= \frac{0.0434}{2} \times (6.65\times 10^{-3})$

$E(1) = 1.44\times 10^{-4} v/m$

E = 1/2 dB/dt (R^2)/r

$E(2)= \frac{0.055^2}{2\times 0.0746} \times (6.65\times 10^{-3})$

$E(2) = 1.34\times 10^{-4} v/m$

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a)

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