Respuesta :
Answer:
[tex]B=15.433\ T[/tex] inwards when viewing from the left side.
Explanation:
Given:
- radius of the circular loop, [tex]r=0.2\ m[/tex]
- current in the coil, [tex]I_c=0.5\ A[/tex]
- Direction of current is clockwise when viewed from left.
- Distance of wire P from the loop, [tex]d_p=0.4\ m[/tex]
- Distance of wire Q from the loop, [tex]d_q=0.8\ m[/tex]
- current in each wires P & Q, [tex]I=0.2\ A[/tex]
Now the magnetic field in coil will be inwards when viewed from left by the Maxwell's right hand thumb rule.
Magnitude is given by:
[tex]B_c=\frac{\mu_0.I_c}{2R}[/tex]
[tex]B_c=\frac{4\pi\times 10^{-7}\times (0.5)}{2\times 0.2}[/tex]
[tex]B_c=15.7\times 10^{-7}\ T[/tex]
Now the effect of magnetic field due to wire P at the center of the loop:
(We get the effective distance as 0.4+0.2=0.6 m)
[tex]B_P=\frac{\mu_0.I}{2\pi.d_p}[/tex]
[tex]B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 0.6}[/tex]
[tex]B_P=6.67\times 10^{-8}\ T[/tex] coming out of the loop when viewed from left.
Now the effect of magnetic field due to wire Q at the center of the loop:
(We get the effective distance as 0.8+0.2=1 m)
[tex]B_P=\frac{\mu_0.I}{2\pi.d_q}[/tex]
[tex]B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 1}[/tex]
[tex]B_P=4\times 10^{-8}\ T[/tex] going in to the loop when viewed from left.
Now the net resultant effect all the magnetic fields:
[tex]B=B_c-B_P+B_Q[/tex]
[tex]B=15.7-0.667+0.4[/tex]
[tex]B=15.433\ T[/tex] inwards when viewing from the left side.
