Question

The conducting triangular loop shown in figure carries a current of 10 A . The magnetic field $\overrightarrow{H}$ at point (0,0,5) due to side 1 of the loop is,

• A. $+65.17{\hat{a}}_y$ (mA/m)
• B. $-59.11{\hat{a}}_y$ (mA/m)
• C. $-65.17{\hat{a}}_y$ (mA/m)
• D. $+79.13{\hat{a}}_y$ (mA/m)

Answer 1

The correct option is B $-59.11{\hat{a}}_y$ (mA/m)

(b)
Note. here that the point of intrest is on z-axis and the side-1 of the triangular is on x-axis . So no need to consider other sides.

The perpendicular from point P on the current filament is at (0,0,0) . The vector and unit vector in the direction of perpendicular towards point P is ,
$\overline{R}=5{\hat{a}}_z$
and ${\hat{a}}_R={\hat{a}}_z$
The angles made by ends of the filament with the perpendicular are (dotted line),
${\alpha }_1=0$ ;
${\alpha }_2$ $=ta{n}^{-1}\left(\frac{2}{5}\right)$
$={21.80}^0$
The current filament along x-axis gives
${\hat{a}}_l={\hat{a}}_x$
Now the direction of H is ,
${\hat{a}}_{\varphi }={\hat{a}}_l\times {\hat{a}}_R$
$={\hat{a}}_x\times {\hat{a}}_z$
$=-{\hat{a}}_y$
The field intensity due to finite filament is,
$\overline{H}=\frac{I}{4\pi R}[\sin {\alpha }_2+\sin {\alpha }_1]{\hat{a}}_{\varphi }$
$=\frac{10}{4\pi \left(5\right)}\left[\sin \right(21.80)+\sin (0\left)\right](-{\hat{a}}_y)$
$\overline{H}=-59.11{\hat{a}}_y\left(mA/m\right)$