24P2Q05

5a)

\displaystyle \begin{aligned}B&={{\mu }_{o}}\frac{\mathrm{I}}{2\pi r} \\   2.8\times {{10}^{-4}}&=(4\pi \times {{10}^{-7}})\frac{\mathrm{I}}{2\pi (8.5\times {{10}^{-3}})} \\  \mathrm{I}&=11.9\text{ A} \end{aligned}

5bi)

\displaystyle \begin{aligned}F&=BIL \\\frac{F}{L}&=BI \\&=(2.8\times {{10}^{-4}})(9.2) \\&=2.58\times {{10}^{-3}}\text{ N }{{\text{m}}^{-1}}  \end{aligned}

5bii)

Weight per unit length\displaystyle =0.063\times 9.81=0.61803\text{ N }{{\text{m}}^{-1}}

Percentage change\displaystyle =\frac{2.58\times {{10}^{-3}}}{0.61803}=0.417%

5biii)

Decrease.

Unlike currents repel. The magnetic force exerted on rod is upward and helps support the weight.

5biv1)

The magnetic field of the wire above the wire is directed out of the page.

The current is upward.

Using Fleming’s Left Hand Rule, the magnetic force on the wire is rightward.

5biv2)

It is inversely proportional to the distance.

COMMENT: This is because the magnetic flux density is inversely proportional to distance.\displaystyle B={{\mu }_{o}}\frac{\mathrm{I}}{2\pi r}

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