Solve B=frac{mu_{0}}{4pi}*I/r^2sinalpha*2pia

$$B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { I } { r ^ { 2 } } \sin \alpha \cdot 2 \pi a$$

Solve for I

$$\left\{\begin{matrix}I=\frac{2Br^{2}}{a\mu _{0}\sin(\alpha )}\text{, }&\mu _{0}\neq 0\text{ and }a\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}\text{ and }r\neq 0\\I\in \mathrm{R}\text{, }&\left(\exists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}\text{ or }\left(r\neq 0\text{ and }\left(a=0\text{ or }\mu _{0}=0\right)\right)\right)\text{ and }B=0\text{ and }r\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by r^{2}.

$$Br^{2}=\frac{1}{4}r^{2}\pi ^{-1}\mu _{0}\times \frac{I}{r^{2}}\sin(\alpha )\times 2\pi a$$

Multiply \frac{1}{4} and 2 to get \frac{1}{2}.

$$Br^{2}=\frac{1}{2}r^{2}\pi ^{-1}\mu _{0}\times \frac{I}{r^{2}}\sin(\alpha )\pi a$$

Multiply \frac{1}{2} times \frac{I}{r^{2}} by multiplying numerator times numerator and denominator times denominator.

$$Br^{2}=\frac{I}{2r^{2}}r^{2}\pi ^{-1}\mu _{0}\sin(\alpha )\pi a$$

Express \frac{I}{2r^{2}}\sin(\alpha ) as a single fraction.

$$Br^{2}=\frac{I\sin(\alpha )}{2r^{2}}r^{2}\pi ^{-1}\mu _{0}\pi a$$

Express \frac{I\sin(\alpha )}{2r^{2}}\pi as a single fraction.

$$Br^{2}=\frac{I\sin(\alpha )\pi }{2r^{2}}r^{2}\pi ^{-1}\mu _{0}a$$

Express \frac{I\sin(\alpha )\pi }{2r^{2}}a as a single fraction.

$$Br^{2}=\frac{I\sin(\alpha )\pi a}{2r^{2}}r^{2}\pi ^{-1}\mu _{0}$$

Express \frac{I\sin(\alpha )\pi a}{2r^{2}}r^{2} as a single fraction.

$$Br^{2}=\frac{I\sin(\alpha )\pi ar^{2}}{2r^{2}}\pi ^{-1}\mu _{0}$$

Cancel out r^{2} in both numerator and denominator.

$$Br^{2}=\frac{\pi Ia\sin(\alpha )}{2}\pi ^{-1}\mu _{0}$$

Express \frac{\pi Ia\sin(\alpha )}{2}\pi ^{-1} as a single fraction.

$$Br^{2}=\frac{\pi Ia\sin(\alpha )\pi ^{-1}}{2}\mu _{0}$$

Express \frac{\pi Ia\sin(\alpha )\pi ^{-1}}{2}\mu _{0} as a single fraction.

$$Br^{2}=\frac{\pi Ia\sin(\alpha )\pi ^{-1}\mu _{0}}{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{\pi Ia\sin(\alpha )\pi ^{-1}\mu _{0}}{2}=Br^{2}$$

Multiply both sides of the equation by 2.

$$\pi Ia\sin(\alpha )\pi ^{-1}\mu _{0}=2Br^{2}$$

The equation is in standard form.

$$a\mu _{0}\sin(\alpha )I=2Br^{2}$$

Divide both sides by a\sin(\alpha )\mu _{0}.

$$\frac{a\mu _{0}\sin(\alpha )I}{a\mu _{0}\sin(\alpha )}=\frac{2Br^{2}}{a\mu _{0}\sin(\alpha )}$$

Dividing by a\sin(\alpha )\mu _{0} undoes the multiplication by a\sin(\alpha )\mu _{0}.

$$I=\frac{2Br^{2}}{a\mu _{0}\sin(\alpha )}$$

Solve for B

$$B=\frac{Ia\mu _{0}\sin(\alpha )}{2r^{2}}$$

$$r\neq 0$$