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sasho [114]
3 years ago
11

A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc

tion: begin mathsize 12px style B with rightwards arrow on top equals 3 space j with logical and on top end style T. The loop carries a 1/2 amp current, I = 0.5 A. If the lengths of the sides of the loop are 4 m and 2 m, what is the magnitude of the magnetic moment, mu, of the loop?
Physics
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

hope this helps!!

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it loses engry  it follows difernt paths

Explanation:

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A wooden cube with the mass of 1kg is placed on a frictionless plane that makes an angle of 30° with the floor.
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2 years ago
A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb
densk [106]
  • Voltage=V=120V
  • Power=P=60W

  • Current=I

\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A

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Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down
MA_775_DIABLO [31]

Answer:

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

          y_total = 5,657 + 12

          y_total = 17,657 m

          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

the distance traveled using the pythagorean theorem is

         r = √ (d1² + d2²)

          r = √ (8² + 12²)

          r = 14.42 m

The time used for this shredding is

         t = t1 + t2

         t = 1 + 1.2

         t = 2.2 s

let's calculate the average speed

         v = 14.42 / 2.2

         v = 6.55 m / s

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