Ask question

Ask question

All categories

2 years ago

6

A wire is formed into a circle having a diameter of 10.0 cm and is placed in a uniform magnetic field of 3.00mT . The wire carri

es a current of 5.00 A. Find (a) the maximum torque on the wire.

1 answer:

mojhsa [17]2 years ago

6 0

The maximum torque on wire is 0.12 x $10^{-3}$ N-m.

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to find the maximum torque on the wire.

<h3>What is the formula to calculate the torque on a wire placed in a uniform magnetic field.</h3>

The torque on a wire placed in a uniform magnetic field is given by -

$$\tau = IAB\;sin\theta$

According to the question, we have -

Diameter of wire (D) = 10 cm

Therefore -

Radius of wire (r) = 5 cm = 1/20 meter.

Magnetic field (B) = 3 mT = 3 x $10^{-3}$ T

I = 5 A

Using the formula for torque -

$$\tau = IAB\;sin\theta$

For maximum torque -

$\theta$ = 90

Therefore -

$$\tau = IAB$ = 5 x 3.14 x $\frac{1}{20}$ x $\frac{1}{20}$ x 3 x $10^{-3}$ = 0.12 x $10^{-3}$ N-m.

Hence, the maximum torque on wire is 0.12 x $10^{-3}$ N-m.

To solve more questions on Torque on wire, visit the link below-

brainly.com/question/13111267

#SPJ4

You might be interested in

You are driving at 23 m/s when you see a red light ahead. Your car is capable of decelerating at a rate of 3.41 m/s2 and it take

Pavel [41]

Answer:

39.2 m

Explanation:

i think

4 0

3 years ago

What is dark energy?

Gnesinka [82]

Dark energy is what is causing the accelerating expansion of the universe. It can also be defined as the theoretical energy that counteracts or acts against gravity. Others also define it as a theoretical, new kind of dynamic energy field that fills space, but whose energy has an effect opposite to that of normal energy.

5 0

3 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

Suppose a distant world with surface gravity of 5.20 m/s2 has an atmospheric pressure of 7.16 ✕ 104 Pa at the surface. (a) What

Nesterboy [21]

Answer:

899752.13598 N

Explanation:

p = Pressure on the plate = $7.16\times 10^4\ Pa$

A = Area = $\pi r^2=\pi 2^2=\pi 4$

F = Force on the region

When force is divided by area we get pressure

$p=\frac{F}{A}\\\Rightarrow F=p\times A\\\Rightarrow F=7.16\times 10^4\times \pi 4\\\Rightarrow F=899752.13598\ N$

Force is exerted by the atmosphere on a disk-shaped region is 899752.13598 N

6 0

3 years ago

What is the simplest relationship between the angular wavenumber k and just one of the other kinematic variables?

Inessa [10]

Answer : $k=\dfrac{\omega}{k}$

Explanation :

The word kinematics means the study of motion. Kinematic variables gives the description of the motion of the body.

Displacement, Velocity, acceleration and time are associated with the motion of particle.

Wavenumber is defined as the frequency of wave, which is measured in cycles per unit distance.

Wave number is also defines as:

$k=\dfrac{2\pi}{\lambda}$

where, $\lambda$ is wavelength

$k=\dfrac{2\pi}{\lambda}\times \dfrac{T}{T}$

since, $\dfrac{2\pi}{T}=\omega \ and\ \dfrac{1}{V}=\dfrac{T}{\lambda}$

So, $k=\dfrac{\omega}{V}$

Where, $\omega$ is angular frequency and $V$ is velocity of wave.

Hence, the relationship between the angular wavenumber k and and kinamatic variable V is

$k=\dfrac{\omega}{V}$

6 0

3 years ago