Ask question

Ask question

All categories

2 years ago

13

Find the magnitude of the magnetic field ∣∣B⃗ (r)∣∣ inside the cylindrical resistor, where r is the distance from the axis of th

e cylinder, in terms of i, r, r0, l, and other given variables. You will also need π and μ0. Ignore fringing effects at the ends of the cylinder.

1 answer:

Elena-2011 [213]2 years ago

6 0

Answer:

The magnetic field inside the cylindrical resistor is $\dfrac{\mu_{0}ir}{2\pir_{0}^2}$

Explanation:

Given that,

Distance from the axis of the cylinder = r

We need to calculate the magnetic field inside the cylindrical resistor

Using formula of magnetic field

$\oint{\vec{B}\cdot\vec{dl}}=\mu_{0}i_{encl}$

$B\cdot(2\pi r)=\mu_{0}\dfrac{i\pir^2}{\pi r_{0}^2}$

Where, r₀ = radius

r = distance

i = current

$|\vec{B}(r)|=\dfrac{\mu_{0}ir}{2\pir_{0}^2}$

Hence, The magnetic field inside the cylindrical resistor is $\dfrac{\mu_{0}ir}{2\pir_{0}^2}$

You might be interested in

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

Whitepunk [10]

It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.

5 0

2 years ago

Read 2 more answers

A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate

liq [111]

Answer:

3 cm

Explanation:

According to the question,

$D=1.5 m$ .

$d=9 cm$ .

$\lambda =0.9 mm$ .

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

$y=2\frac{\lambda (D)}{d}$ .

Substitute all the variable in above equation.

$y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}$ .

$y=3 cm$ .

5 0

2 years ago

A FBD of a rocket launching into space should include:

Vladimir [108]

Answer:

Explanation:

the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.

3 0

1 year ago

An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t

Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

$\omega = positive$

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

$\alpha = negative$

so here correct answer is

B. W is positive and a is negative

8 0

2 years ago

Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi

aleksklad [387]

1) The buoyant force acting on an object immersed in a fluid is:
$B=d_f V_d g$
where $d_f$ is the density of the fluid, $V_d$ is the volume of displaced fluid, and $g=9.81~m/s^2$ is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is $d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3$ . Using the relationship between density, volume and mass, we find
$V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3$

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is $d_f = 1~Kg/m^3$ , so we have
$B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N$

4) The weight of 1 Kg of titanium is instead:
$W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N$
So, the buoyant force is negligible compared to the weight.

7 0

1 year ago