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3 years ago

What is a magnetometer and how does it work?

1 answer:

8 0

Magnetometers<span> are widely used for measuring the Earth's magnetic field and in geophysical surveys to detect magnetic anomalies of various types. </span>They<span> are also used in the military to detect submarines.</span>

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High exposure to which type of electromagnetic radiation can cause genetic

melamori03 [73]

The answer is microwaves

3 0

3 years ago

Read 2 more answers

A circular loop of wire 75 mm in radius carries a current of 113 A. Find the (a) magnetic field strength and (b) energy density

Roman55 [17]

The magnetic field strength is 9.47 ×10⁻⁴ T

The energy density at the center of the loop is 0.36 J/m³

<h3>Calculating Magnetic field strength & Energy density </h3>

From the question, we are to find the magnetic field strength

The magnetic field strength of a loop can be calculated by using the formula,

$B = \frac{\mu_{0} I}{2R}$

Where B is the magnetic field strength

$\mu_{0}$ is the permeability of free space $(\mu_{0}=4\pi \times 10^{-7} \ N/A^{2})$

$I$ is the current

and R is the radius

From the give information,

$R = 75 \ mm= 75 \times 10^{-3} \ m$

and $I = 113 \ A$

Putting the parameters into the formula, we get

$B = \frac{4\pi \times 10^{-7} \times 113}{2 \times 75 \times 10^{-3} }$

$B = 9.47 \times 10^{-4} \ T$

Hence, the magnetic field strength is 9.47 ×10⁻⁴ T

Now, for the energy density

Energy density can be calculated by using the formula,

$u_{B} = \frac{B^{2} }{2\mu_{0} }$

Where $u_{B}$ is the energy density

Then,

$u_{B}= \frac{(9.47\times 10^{-4} )^{2} }{2 \times 4\pi \times 10^{-7} }$

$u_{B} = 0.36 \ J/m^{3}$

Hence, the energy density at the center of the loop is 0.36 J/m³

Learn more on Magnetic field stregth & Energy density here: brainly.com/question/13035557

7 0

2 years ago

A physicist is constructing a solenoid. She has a roll of insulated copper wire and a power supply. She winds a single layer of

Leni [432]

Answer:

P =105.44 W

Explanation:

Given that

D= 10 cm ,L= 60 cm

d= 0.1 cm ,B= 6.4 mT

ρ= 1.7 x 10⁻⁸ Ω · m

The number of turns N

N= L/d

N= 60/0.1 = 600 turns

Length of wire

Lc= πDN

Lc= 3.14 x 0.1 x 600

Lc=188.4 m

The magnetic filed given as

$B=\dfrac{\mu_oNI}{L}$

$I=\dfrac{BL}{\mu_oN}$

Now by putting the values

$I=\dfrac{0.0064\times 0.6}{4\pi \times 10^{-7}\times 600}$

I=5.09 A

The resistance R given as

$R=\dfrac{\rho L_c}{A}$

$A=\dfrac{\pi}{4}\times d^2$

$A=\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2$

$R=\dfrac{1.7\times 10^{-8} \times 188.4}{\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2}$

R=4.07 Ω

Power P

p =I²R

P= 5.09² x 4.07 W

P =105.44 W

5 0

3 years ago

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

zhannawk [14.2K]

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}$

$v = 16.52 m/s$

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}$

$v = 7.47 m/s$

4 0

3 years ago

What speed do all electromagnet waves travel

Grace [21]

3x10^8 m/s, or the ‘speed of light’ - this applies to the entire spectrum

3 0

2 years ago