What is its moment of inertia about an axis that passes through its center?

A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A

Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

$F = ILBsin \theta$

Here,

I = Current

L = Length

B = Magnetic Field

$\theta$ = Angle between Force and Magnetic Field

But $\theta = 90\°$

$F = ILB$

Rearranging to find the Magnetic Field,

$B = \frac{F}{IL}$

Here the force per unit length,

$B = \frac{1}{I}\frac{F}{L}$

Replacing with our values,

$B = \frac{0.130N/m}{16.5}$

$B = 0.0078T$

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0

3 years ago

If you place a paper clip very close to a magnet, the paper clip

Alik [6]

The answer is “D. all of the above”!

Metal from the paper clip is attracted to the magnet, so it will naturally move toward and stick to the magnet. This will cause the paper clip to temporarily become a magnet for other metals. I hope this helped!

7 0

1 year ago

Visible electrical discharge from a storm cloud is called?

SOVA2 [1]

id have to say its thunder

7 0

3 years ago

What is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)<br> please someone help me

777dan777 [17]

Answer:

60mph=26.8224meters per second

Explanation:

4 0

3 years ago

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$