Question

1. Researchers have tracked the head and body movements of several flying insects, including blowflies, hover flies, and honeybees. They attach lightweight, flexible wires to a small metal coil on the insect's head, and another on its thorax, and then allow it to fly in a stationary magnetic field. As the coils move through the field, they experience induced emfs that can be analyzed by computer to determine the corresponding orientation of the head and thorax. Suppose the fly turns through an angle of 90 ∘ in 36 ms . The coil has 99 turns of wire and a diameter of 2.0 mm . The fly is immersed in a magnetic field of magnitude 0.13 mT .
If the magnetic flux through one of the coils on the insect goes from a maximum to zero during this maneuver find the magnitude of the induced emf.

2. You would like to construct a 46.6 mH inductor by wrapping insulated copper wire (diameter = 3.29×10−2 cm ) onto a tube with a circular cross section of radius 2.53 cm .

What length of wire is required if it is wrapped onto the tube in a single, close-packed layer?

Answer 1

Answer:

a) $E = 1.12 * 10^{-6} V$

b) Length of the wire, l = 2.01 m

Explanation:

Number of turns, N = 99

diameter of the coil, d = 2.0 mm = 0.002 m

Area of the coil:

$A = \frac{\pi d^{2} }{4} \\A = \frac{\pi 0.002^{2} }{4} \\A = 0.00000314 m^{2}$

Time interval, t = 36 ms = 0.036 s

The fly turns through an angle of 90°

Magnitude of the magnetic field, B = 0.13 mT = 0.00013 T

From Faraday's law of Electromagnetism:

$E = -N\frac{\triangle \phi}{\triangle t}$

$\triangle \phi = BA [cos\phi_{2}) - cos\phi_{1}]\\\triangle \phi = (0.00013*0.00000314) [cos90 - cos0]\\ \triangle \phi = - 408.2 * 10^{-12} wb$

$E = -99\frac{-408.2 * 10^-12}{0.036} \\E = 1.12 * 10^{-6} V$

2) The inductance of the inductor, L = 46.6 mH = 0.0466 H

Diameter of the copper wire, d = 3.29 * 10⁻² cm = 0.000329 m

Cross sectional radius of the tube, R = 2.53 cm = 0.0253 m

Area, A = πR² = π *0.0253² = 0.002 m²

For a closely packed layer, l = Nd

N = l/d

The inductance of the inductor can be given by the formula :

$L = \frac{\mu N^{2} A}{l} \\N = l/d\\L = \frac{\mu l^{2} A}{d^{2} l}\\L = \frac{\mu lA}{d^{2} }\\l = \frac{d^{2}L}{\mu A }\\l = \frac{0.000329^{2}*0.0466}{4\pi*10^{-7} 0.002 }$

Length of the wire, l = 2.01 m