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OleMash [197]
3 years ago
10

I) A circular coil with radius 20 cm is placed with it's plane parallel and between two straight

Physics
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

B=15.433\ T inwards when viewing from the left side.

Explanation:

Given:

  • radius of the circular loop, r=0.2\ m
  • current in the coil, I_c=0.5\ A
  • Direction of current is clockwise when viewed from left.
  • Distance of wire P from the loop, d_p=0.4\ m
  • Distance of wire Q from the loop, d_q=0.8\ m
  • current in each wires P & Q, I=0.2\ A

Now the magnetic field in coil will be inwards when viewed from left by the Maxwell's right hand thumb rule.

<u>Magnitude is given by:</u>

B_c=\frac{\mu_0.I_c}{2R}

B_c=\frac{4\pi\times 10^{-7}\times (0.5)}{2\times 0.2}

B_c=15.7\times 10^{-7}\ T

<u>Now the effect of magnetic field due to wire P at the center of the loop:</u>

(We get the effective distance as 0.4+0.2=0.6 m)

B_P=\frac{\mu_0.I}{2\pi.d_p}

B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 0.6}

B_P=6.67\times 10^{-8}\ T coming out of the loop when viewed from left.

<u>Now the effect of magnetic field due to wire Q at the center of the loop:</u>

(We get the effective distance as 0.8+0.2=1 m)

B_P=\frac{\mu_0.I}{2\pi.d_q}

B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 1}

B_P=4\times 10^{-8}\ T going in to the loop when viewed from left.

<u>Now the net resultant effect all the magnetic fields:</u>

B=B_c-B_P+B_Q

B=15.7-0.667+0.4

B=15.433\ T inwards when viewing from the left side.

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