
Here $\mathrm{M}=2 \mathrm{n}(2 \mathrm{I})(2 \mathrm{a})$
$M=8$ $nI \mathrm{a}$
$\therefore \quad \tau=M B \sin \left(90^{\circ}-60^{\circ}\right)$
$\tau=M B \cos 60^{\circ}$
$\tau=8$ nIa $\cos 60^{\circ}$
Reason : Resistance of a voltmeter is very large.
$(i)\,\,\left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{d\vec l\, \times \,\vec r}}{{{r^3}}}} \right)$
$(ii)\,\, - \left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{d\vec l\, \times \,\vec r}}{{{r^3}}}} \right)$
$(iii)\,\left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{\,\vec r \times d\vec l}}{{{r^3}}}} \right)$
$(iv)\, - \left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{\,\vec r \times d\vec l}}{{{r^3}}}} \right)$
