Question

A wheel with 4 spokes is placed with its plane perpendicular to a uniform magnetic field B of magnitude 0.5 T. The field is directed into the plane of the paper and is present over the entire region of the wheel as shown in fig. When the switch S is closed, there is initial current of 6 A between the axis and the rim and the wheel begins to rotate. Resistances of the spokes are 1, 2, 4 and $8\Omega$, respectively. Resistance of rim is negligible. Radius of the wheel is 0.2 m. The initial torque acting on the wheel is $6\times {10}^{-x}Nm$. Find x than.

Answer 1

$r=0.2m$
Suppose the torque acting on the spokes with resistances 1, 2, 4 and $8\Omega$ are ${\tau }_1,{\tau }_2,{\tau }_3$ and ${\tau }_4$, respectively.
Then, as all these torques act in the same direction
${\tau }_{net}={\tau }_1+{\tau }_2+{\tau }_3+{\tau }_4$
$\Rightarrow \tau =i_{1}Br\frac{r}{2}+i_{2}Br\frac{r}{2}+i_{4}Br\frac{r}{2}+i_{8}Br\frac{r}{2}$
$=\frac{B{r}^2}{2}(i_1+i_2+i_4+i_8)=\frac{iB{r}^2}{2}$
$\Rightarrow \tau =\frac{6\times 0.5\times (0.2{)}^2}{2}=1.5\times 0.04=60\times {10}^{-3}$
$=0.06Nm$
Hence, $\tau =0.06Nm\odot$