Question

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

Answer 1

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.

Magnitude is given by:

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

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

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

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)

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

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

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

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)

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

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

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

Now the net resultant effect all the magnetic fields:

$B=B_c-B_P+B_Q$

$B=15.7-0.667+0.4$

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