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sasho [114]
4 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]4 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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tatiyna
<span>The formulas are, Impulse = mv-mu ....... (1) v^2 = u^2 + 2as .......... (2) We know that, u=0 a=acceleration=gravity = 9.80665 m/s^2 = 9.81 m/s^2 s=19.6 sub (2) we get, v^2 = 0+ 2*9.81*19.6 v^2 = 2*9.81*19.6 v^2 = 384.552 v = 19.6099 v = 19.61 m/s Sub v=19.61 m/s in (1) we get, Impulse = mv - mu we know that u=0; v= 19.61 m/s; m= 3.00 kg Impulse = 3(19.61) - 3(0) Impulse = 58.83-0 Impulse = 58.83 Ns. Therefore the gravitational force exerted by the stone is 58.83 Ns.</span>
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4 years ago
Why is the formation of hydrogen bonds between water molecules categorized as cohesion
marishachu [46]

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The oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Then, it can be said that a water molecule will have a negative side (oxygen) and a positive side (hydrogen).

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Answer:

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