Figure 28-46 shows a wood cylinder of mass $\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{\text{kg}}$ and length$\mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{\text{m}}$,with$\mathbf{N}\mathbf{}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{0}$ turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long central axis of the cylinder. The cylinder is released on a plane inclined at an angle $\mathit{\theta }$ to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude$\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{\text{T}}$ , what is the least current ithrough the coil that keeps the cylinder from rolling down the plane?

• 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" ${\tau }_{net}=I\alpha$

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

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

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

${\tau }_{net}=I\alpha$

Applying this equation and substituting$\alpha =0$for equilibrium condition, we get

$f\times r-NiAB\text{sin}\theta =0······\left(2\right)$

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

$A=L\times \left(2r\right)=2rL$

Using this in the above equation, we get

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

Rearranging this equation for current

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

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