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

Answer 2

Faraday's law allows finding the results for the questions about insect movement and inductance are:

      1) The indicated electromotive force is: emf = 1.12 10⁻⁶ V

     2) The length of the wire is: l = 2.01 m

Faraday's law of induction states that the induced electromotive force in a circuit is proportional to the change in magnetic flux.

 

          fem = $-N \ \frac{d \Phi_B }{dt}$  

the magnetic flux is

         $\Phi_B$ = B . A = B A cos θ

where B is the magnetic field, A the normal vector to the area, θ the angle between them, N the number of turns and t the time

They indicate that the field and area of ​​the loop are constant.

          fem = $- N B A \ \frac{d cos \theta }{dt }$  

They indicate the turn number N = 99, the diameter of the loop d = 2.0 mm= 2.0 10⁻³ m and the time it takes for the change in position Δt = 36ms = 36 10⁻³ s. We substitute the derivative for the variation.

           

            $fem = -NBS \ \frac{\Delta cos \theta }{\Delta t}$

The area of ​​the turn is

         A = π r² =πi d²/4

The magnetic field is of constant magnitude B = 0.13 mT = 0.13 10⁻³ T, we substitute Faraday's law.

       $fem = - NB \pi \frac{d^2}{4} \ \frac{cos \frac{\pi }{2} - cos 0}{\Delta t}$  

Let's calculate.

     $fem= \frac{99 \ 0.13 \ 10^{-3} \pi \ (2.0 \ 10^{-3})^2 }{36 \ 10^{-3} }$  

      fem = 1.12 10⁻⁶ V

2) The inductance is the resistance of the cable to the change in current, it is obtained from Faraday's law.

           L = $N \frac{\Phi_B}{I}$  

where N is the number of the turn, Ф_B the magnetic flux and I the current.

The magnetic field inside the inductor is

       

          B = $\mu_o \frac{N}{l} \ I$  

We look for the magnetic flux.

           Ф_B  = B. A = B A cos θ

in a solenoid the magnetic field and the normal one have the same direction, so the angle is zero.

          $\Phi_B = \mu_o \frac{N}{l} \ \pi r^2$

Let's substitute into induction.

       $L = \mu_o \pi \ \frac{N^2 \ r^2 }{l}$  

The length of the solenoid is:

        $l = N d \\N = \frac{l}{d}$  

where l is the length of the solenoid and d the diameter of the wire, we substitute.

       $L = \mu_o \pi \frac{l^2 r^2}{d^2 l}$  

       $L = \mu_o \pi \ \frac{l\ r^2}{d^2 }$  

       $l = \frac{L \ d^2 }{\mu_o \pi r^2 }$  

Let's calculate

       $l = \frac{46.6 \ 10^{-3} 3.29 \ 10^{-2} }{4 \pi \ 10^{-7} (2.53 \ 10^{-2})^2 }$  

      l = 2.01 m

In conclusion using Faraday's law we can find the results for the questions about insect motion and inductance are:  

      1) The indicated electromotive force is: fem = 1.12 10⁻⁶ V

      2) The length of the wire is: l = 2.01 m

Learn more about Faraday's law here:  brainly.com/question/1640558