Question # 3

A closely wound, circular coil with radius 2.80 cm has 720 turns. Part APart complete What must the current in the coil be if th

lara [203]

Answer:

I = 4.642 Ampere.

x = 2.145 cm

Explanation:

a) As we know, the magnetic field on the axis of the loop is given as

$B = \frac{ \mu NIa^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }$

where a = radius of loop

x = point on the axis of loop

N = No of turns of coil

Current in the loop for which the magnetic field at the center is 0.0750 Tesla is given as x = 0

Therefore, the above equation can be rewritten as

$B_{x} = \frac{ \mu NI}{2a}$

I = $\frac{2aB}{μN}$

by putting values

I = $\frac{2 X 0.0750 T X 0.028 m}{4\pi X 10^{-7} T.\frac{m}{A} X 720}$

Current in the loop for which the magnetic field at the center is 0.0750 Tesla = I = 4.642 Ampere

b)

Now for part b the magnetic field at a distance x from the center is given as

$B = \frac{ \mu N I a^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }$

multiply and divide by a on both sides we get

$B = \frac{ \mu NI}{2a} X \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$

As we know, according to Biot sivorts law,the Magnetic Field at the Center of a circular loop is given as

$B = \frac{ \mu NI}{2a}$ ( Magnetic field at center) = $B_{c}$

So we got magnetic field at any point x as

$B_{x}$ = $B_{c}$ X $\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$

For magnetic field at x is half of the B at center

$B_{x}$ = $\frac{1}{2}$ $B_{c}$

from the above two equations

$\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$ = $\frac{1}{2}$

$(x^{2} + a^{2})^{3} = 4a^{6}$

$x = \sqrt{4^{\frac{1}{3}} -1 } X a^{}$

Putting a = 2.80 cm

We have x = 2.145 cm Ans

μ

6 0

3 years ago

4. A banking system provides users with several services:

VLD [36.1K]

A diagram showing a use case diagrams for these requirements is given in the image attached.

<h3>What is system Case diagram?</h3>

A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.

A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.

Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.

Learn more about Case diagram from

brainly.com/question/12975184

#SPJ1

4 0

2 years ago

What are three reasons why you should make your first prototype on a smaller scale?

erastovalidia [21]

Here’s a few -
Save time.
Save money.
Understand the problem better.
Get feedback quickly.
Procrastinate less and get started quickly.
Build a better product through many iterations.

Hope this helps !

8 0

3 years ago

Read 2 more answers

A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur

Lina20 [59]

Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and increasing the capacitance proportionally to the dielectric constant of the Teflon.

8 0

3 years ago

(2pts) In _______________, the pulse amplitude is made proportional to the amplitude of the modulating signal. pulse-amplitude m

Reptile [31]

Answer:

(a) pulse-amplitude modulation (PAM)

Explanation:

It should be no surprise that pulse amplitude is modulated in <em>pulse-amplitude modulation (PAM)</em>.

7 0

3 years ago