Question 272: Magnetism and Matter

Question

A loop ABCDA , carrying current $\mathrm{I}=12 \mathrm{~A}$, is placed in a plane, consists of two semi-circular segments of radius $R_1=6 \pi \mathrm{~m}$ and $\mathrm{R}_2=4 \pi \mathrm{~m}$. The magnitude of the resultant magnetic field at center O is $\mathrm{k} 	imes 10^{-7} \mathrm{~T}$. The value of k is_________.
( Given $\mu_0=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$ )

Accepted Answer

Magnetic Field from Semi-Circular Segments
:
The magnetic field at the center due to a semi-circular loop is given by:
$      B = \frac{\mu_0 I}{4R}      $
Calculating $B_{R_1}$ and $B_{R_2}$
:
For the semi-circle with radius $R_1$:
$      B_{R_1} = \frac{\mu_0 I}{4R_1} = \frac{4\pi 	imes 10^{-7} 	imes 12}{4 	imes 6\pi}      $
For the semi-circle with radius $R_2$:
$      B_{R_2} = \frac{\mu_0 I}{4R_2} = \frac{4\pi 	imes 10^{-7} 	imes 12}{4 	imes 4\pi}      $
Net Magnetic Field at the Center (O)
:
The total magnetic field $B_0$ at the center is the difference between these two fields (since they are in opposite directions):
$      B_0 = |B_{R_1} - B_{R_2}|      $
Calculate:
$      B_0 = \frac{4\pi 	imes 10^{-7} 	imes 12}{4} \left(\frac{1}{4\pi} - \frac{1}{6\pi}ight)      $
$      B_0 = 12\pi 	imes 10^{-7} \left(\frac{1}{12\pi}ight)      $
$      B_0 = 1 	imes 10^{-7} 	ext{ T}      $
Value of $k$
:
Given that the magnitude of the resultant magnetic field is $k 	imes 10^{-7} 	ext{ T}$, we find $k = 1$.
Thus, the value of $k$ is 1.

Magnetism and Matter

Integer / Numeric Answer Type