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yarga [219]
3 years ago
13

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.
Physics
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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