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Nadusha1986 [10]

3 years ago

13

Two concentric current loops lie in the same plane. The smaller loop has a radius of 2.7cm and a current of 12 A. The bigger loo

p has a current of 20 A. The magnetic field at the center of the loops is found to be zero. What is the radius of the bigger loop?

2 answers:

Kryger [21]3 years ago

8 0

Answer:

Radius of bigger loop(R) =4.5cm

Explanation:

Consider a circular path of radius r around the wire. The magnetic field along that path is given by ;

∫B*dl = k*I where I is the current enclosed. From symmetry, ∫B*dl = 2*π*r*B

B = K*I/r, so the magnetic field varies inversely as the loop radius and directly as the current.

The smaller loop current to radius ratio is 12/2.7

The bigger loop current to radius ratio is = 20/R

12/2.7 = 20/R

R = (20 * 2.7)/12

R=54/12

R=4.5cm

Dmitry [639]3 years ago

3 0

Answer:

Explanation:

The B-field at the center of a circular loop of radius, r and current, I is;

Magnetic field, B = (μ × I) ÷ 2pi × R

Given:

rs = 2.7 cm

= 0.027 m

Is = 12 A

Ib = 20 A

(μ × Is)/2 × rs = (μ × Ib)/2 × rb

Inputting values,

rb = (20 × 2 × 0.027)/12 × 2

= 0.045 m

= 4.5 cm.

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a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

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Explanation:

The spacetime interval $\Delta s^2$ is given by

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please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

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$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

<h3>b.</h3>

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$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

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$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

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$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

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$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

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