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2 years ago

Why does a torque rod have a stronger magnetic field than a torque coil?

1 answer:

8 0

Magnetic torquer or torque rod are essentially sets of electromagnets that are laid out to create a magnetic field that interfaces with an ambient magnetic field, so that the counter-forces produced is a stronger magnetic force. In addition, its magnetic field strength increases with the current due to the rod that has field lines pointing in the same direction, closed together and nearly parallel with uniformly spaced. This set up indicates that the field of a torque rod produced powerful magnetic field than a torque coil.

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

A baseball thrown from the outfield to home plate does not have which of the following types of energy while it's in the air? A.

posledela

The ball does not have D. Radiant energy.

7 0

2 years ago

A 500kg car is driving through a neighborhood at 20m/s. Then the car hits a speed bump which causes it to slow to 15m/s. What wa

lawyer [7]

Based on the formula for calculating impulse, the impulse of the speed bump to the car is 2500 Ns.

<h3>What is the impulse of the speed bump?</h3>

Impulse = change in momentum

momentum = mass * velocity

Change in momentum = mu - mv

where u is initial velocity

v is final velocity

Impulse = 500 * 20 - 500 * 15

Impulse = 2500 Ns

Therefore, the impulse of the speed bump to the car is 2500 Ns.

Learn more about impulse at: brainly.com/question/297527

5 0

2 years ago

The two forces in each pair can either both act on the same body or they can act on different bodies. The two forces in each pai

Marrrta [24]

Answer:

The correct answer in which peer forces each act in a different body

Explanation:

Newton's law of action and reaction states that the two forces have equal magnitude, but in the opposite direction, each acting on a body. Which creates acceleration in these, if they act in the same body it would be canceled so there would be no movement.

The correct answer in which peer forces each act in a different body

6 0

3 years ago

i)Distinguish between different methods of charging. ii) You are provided with a positively charged gold leaf electroscope. Stat

AleksAgata [21]

Answer:

Explanation:

On rubbing a glass rod with silk, the electrons from the glass rod get transferred to the silk. The silk now has an excess of electrons and so is negatively-charged. On the other hand, the glass rod is deficient in electrons and hence is positively-charged.

In the above case, the silk undergoes negative electrification.

Now, when the positively charged glass rod is touched on the disc of a negatively charged gold leaf electroscope, the electrons shifts towards rod, hence amount of charge on gold leaves decreases and the divergence between the gold leaves decreases as unlike charges attract each other.

Hence, the divergence decreases when a glass rod rubbed with silk is brought near the disc of negatively charged electroscope.

hope it helps pls mark me as brainliest

6 0

2 years ago