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horsena [70]
3 years ago
10

A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a lon

g, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).
Physics
1 answer:
romanna [79]3 years ago
5 0

Answer:

I_{1} = (πDI_{2})/R

Explanation:

If we define the magnitude of the field as B, then we have:

Total magnitude of the field B_{t} = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.

B_loop = (μ0)*(I_{2})/2*R

B_wire = (μ0)*(I_{1})/2*π*D

Thus:

B_loop = B_wire at the center of the loop.

(μ0)*(I_{2})/2*R = (μ0)*(I_{1})/2*π*D

I_{1} = (πDI_{2})/R

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Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23
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Answer:

The ratio is   \frac{T_1}{T_2}  = 3.965

Explanation:

From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  R_1  =  23500 \  km

Generally from Kepler's third law

    T^2 =  \frac{ 4 *  \pi^2 *  R^3}{G * M  }

Here M is the mass of Mars which is constant

        G is the gravitational  constant

So we see that \frac{ 4 *  \pi^2  }{G * M  } =  constant

   

    T^2 = R^3   *  constant      

=>  [\frac{T_1}{T_2} ]^2 =  [\frac{R_1}{R_2} ]^3

Here T_1 is the period of Deimos

and  T_1 is the period of  Phobos

So

      [\frac{T_1}{T_2} ] =  [\frac{R_1}{R_2} ]^{\frac{3}{2}}

=>    \frac{T_1}{T_2}  =  [\frac{23500 }{9380} ]^{\frac{3}{2}}]

=>    \frac{T_1}{T_2}  = 3.965

   

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