Ask question

Ask question

All categories

3 years ago

12

A circular coil that has N = 180 N=180 turns and a radius of r = 13.0 cm r=13.0 cm lies in a magnetic field that has a magnitude

of B 0 = 0.0790 T B0=0.0790 T directed perpendicular to the plane of the coil. What is the magnitude of the magnetic flux Φ B ΦB through the coil?

1 answer:

Tanzania [10]3 years ago

8 0

Answer:

The magnitude of the magnetic flux through the coil is $0.75\ T-m^2$ .

Explanation:

Number of turns in a circular coil, N = 180

Radius of the coil, r = 13 cm = 0.13 m

Magnitude of magnetic field, B = 0.079 T

The magnetic field is directed perpendicular to the plane of the coil. We need to find the magnitude of the magnetic flux through the coil. The magnetic flux is given by the formula as :

$\phi=NBA\ \cos\theta$

$\theta=0$

$\phi=180\times 0.079\times \pi (0.13)^2\\\\\phi=0.75\ T-m^2$

So, magnitude of the magnetic flux through the coil is $0.75\ T-m^2$ . Hence, this is the required solution.

You might be interested in

How does a second class lever make our work easier

dmitriy555 [2]

Answer:

In a second class lever, the load is located between the effort and the fulcrum. If the load is closer to the fulcrum than the effort, then less effort will be required to move the load. If the load is closer to the effort than the fulcrum, then more effort will be required to move the load.

Explanation:

from what i learned, if its farther away from the load, its easier to lift, like a wheel barrel

7 0

2 years ago

What voltage would be measured across the 45 ohm resistor?

Tamiku [17]

You would find the potential difference aka voltage, but more specifically it would be just the voltage that the resistor uses and not the whole circuit.

But if you want the voltage value it’s V=IR so whatever the current is multiply it by the 45 ohm resistor value

8 0

3 years ago

Read 2 more answers

A car is traveling at a speed of 50 m/sec when it suddenly accelerates at a rate of 5 m/sec2. How fast will the car be going (fi

NISA [10]

Answer:

200 ms-1

Explanation:

v = u+ at

v = 50+ 5×30

v = 200 ms-1

5 0

2 years ago

The current running through a toaster oven is 7.5 Amperes when it is connected to 120 volts of potential difference. What is the

Dafna11 [192]

Power rating = volts x amps

Power rating = 7.5 x 120 = 900 watts

Answer: 900 watts

3 0

2 years ago

You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen

Paladinen [302]

Answer:

(A)

$\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0$

<em>(B) D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2D}{g}}$

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

$\displaystyle t=\frac{D}{v_s}$

(A)

The total measured time is the sum of both times and it's given as $t_t=3.5\ seconds$

$\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}$

From this equation we'll manage to compute D

First, we isolate the square root

$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

$\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2$

Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

Now, let's put in numbers:

$g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec$

$\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0$

Computing all the coefficients:

$\displaystyle D^2-26,705.82D+1,458,056.25=0$

Solving for D, we have two possible solutions:

$D=54.71,\ D=26,651.11$

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

$\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec$

$\displaystyle t_2=\frac{54.71}{345}=0.16\ sec$

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0

3 years ago