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1 year ago

What is the direction of the magnetic field b⃗ net at point a? Recall that the currents in the two wires have equal magnitudes.

Physics

1 answer:

andrew11 [14]1 year ago

6 0

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that the currents in the two wires have equal magnitudes, Then:

$\bar{B_{net}} = \bar{B_1} + \bar{B_2}$

$\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}$

$\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out$

Thus; $\bar{B_{net}}$ points out of the screen at A.

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A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

= 3/2×(5.3)×(8.31451)×(24 + 273)

= 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0

1 year ago

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How

tensa zangetsu [6.8K]

Answer:

The number of turns is $N = 1750 \ turns$

Explanation:

From the question we are told that

The inner radius is $r_i = 12.0 \ cm = 0.12 \ m$

The outer radius is $r_o = 15.0 \ cm = 0.15 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic field is $B = 3.75 mT = 3.75 *10^{-3} \ T$

The distance from the center is $d = 14.0 \ cm = 0.14 \ m$

Generally the number of turns is mathematically represented as

$N = \frac{2 * \pi * d * B}{ \mu_o * r_o }$

Generally $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

$N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }$

$N = 1750 \ turns$

5 0

1 year ago

An electric dipole with dipole moment p⃗ is in a uniform electric field E⃗ . A. Find all the orientation angles of the dipole me

Tasya [4]

Answer:

Explanation:

A) When a dipole is placed in an electric field , it experiences a torque equal to the following

torque = p x E = p E sinθ , where θ is angle between direction of p and E .

It will be zero if θ = 0

or if both p and E are oriented in the same direction.

It is the stable orientation of dipole.

If θ = 180° ,

Torque = 0

In this case both p and E are oriented in opposite direction .

It is the unstable orientation of the dipole because if we deflect the dipole by even small angle , it goes back to most stable orientation due to torque acting on it by electric field.

3 0

2 years ago

Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.

marishachu [46]

Answer:

$h=\frac{1}{2}\frac{v^2}{g}$

Explanation:

Let's assume that an object is launched straight upward in a gravitational field. Its initial kinetic energy is given by

$K=\frac{1}{2}mv^2$ (1)

where m is the mass and v is the initial speed.

As the object goes higher, its kinetic energy decreases and it is converted into gravitational potential energy, since the total mechanical energy (sum of kinetic and potential energy) must remain constant:

$E=K+U=const.$

At the highest point of the trajectory, the speed of the object is zero (v=0), so the kinetic energy is also zero (K=0), which means that all the kinetic energy has been converted into potential energy:

$U=mgh$ (2)

where g is the gravitational acceleration and h is the maximum height of the object.

Due to conservation of energy, we can write that (1) and (2) are equal, so:

$\frac{1}{2}mv^2=mgh$

from which we can derive an expression for the maximum height reached by the object

$h=\frac{1}{2}\frac{v^2}{g}$

5 0

1 year ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed