91Ó°ÊÓ

• Number of turns, N= 10.0
• The mass of cylinder, $m=0.250\text{kg}$
• The length of cylinder,L = 0.100 m
• Magnetic field, B = 0.500 T

We use the equations of equilibrium of force as well as torque as the rolling motion is the combination of translational motion and rotational motion.

Formulae:

$F_{net}=ma$

role="math" localid="1662532505010" ${\mathrm{Ï„}}_{net}=I\mathrm{Î\pm }$

The equation of Newton’s second law of motion is

$F_{net}=ma$

Applying this equation for the direction along inclined plane and substituting a = 0 for equilibrium condition, we get

$f-mg\text{sin}\mathrm{θ}=0······\left(1\right)$

Now, the equation of Newton’s second law of motion for rotational scenario is

${\mathrm{Ï„}}_{net}=I\mathrm{Î\pm }$

Applying this equation and substituting$\mathrm{Î\pm }=0$for equilibrium condition, we get

$f×r-NiAB\text{sin}\mathrm{θ}=0······\left(2\right)$

Solving equation (1) and (2) we get

$mgr=NiAB$

Now, the cylinder is rectangular in shape with length L and width (2r), so its area will become

$A=L×\left(2r\right)=2rL$

Using this in the above equation, we get

$mgr=Ni\left(2rL\right)B$

Rearranging this equation for current

$$\begin{gathered} i=\frac{mg}{2NLB} \\ i=\frac{0.250×9.81}{2×10.0×0.100×0.500} \\ i=2.45\text{A} \end{gathered}$$

Hence, the minimum current is $i=2.45\text{A}$