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antoniya [11.8K]
1 year ago
8

A current of 17.0 mA is maintained in a single circular loop of 2.00 \mathrm{~m} circumference. A magnetic field of 0.800T is di

rected parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop.
Physics
1 answer:
poizon [28]1 year ago
3 0

The magnetic moment is -

M = 5.5 x 10^{-3} Am^{2}.

We have current carrying single circular loop placed in a magnetic field

parallel to the plane of the loop.

We have to determine the Magnetic moment of the loop.

<h3>What is the formula to calculate the magnetic moment of loop?</h3>

The formula to calculate the magnetic moment of the loop is -

M = NIA

where -

N -  Number of turns

I - Current in the loop

A - Area of the loop

According to the question, we have -

I = 17mA = 0.017A

Circumference (C) = 2 meters

B = 0.8 T

Now -

C = 2

2\pir = 2

\pir = 1

r = \frac{1}{\pi }

r = 0.32

Using the formula -

M = NIA

M = \piNIr^{2}

M = 3.14 x 1 x 0.017 x 0.32 x 0.32

M = 3.14 x 4 x 0.017

M = 5.5 x 10^{-3} Am^{2}

Hence, the magnetic moment is -

M = 5.5 x 10^{-3} Am^{2}

To solve more questions on Magnetic moments, visit the link below-

brainly.com/question/13933686

#SPJ4

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Explanation:

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<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

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Sum of forces at point B in the y direction:

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3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

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∑F = ma

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∑F = ma

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∑τ = Iα

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3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

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3/√13 Fde = (1.7 kg) (10 m/s²)

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