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kotykmax [81]
1 year ago
6

A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de

termine the magnitude of the magnetic field at the center of the loop.
Physics
1 answer:
docker41 [41]1 year ago
8 0

Answer:

Magnetic field at the center of the loop B=5.89\times 10^{-5}\ T.

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.

We know , magnetic field at the center of loop is given by :

B=N\dfrac{\mu_o i}{2r}

Putting all values in above equation we get :

B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.

Hence , this is the required solution.

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0.1  

Explanation:

a) From the restoring Force we know that :  

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the gravitational force :  

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Where:

F_r is the restoring force .

F_g is the gravitational force

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k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

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