Question

$\frac{d^2B_x}{dy^2} = -3\frac{\mu_0I}{2}Na^{-1}\frac{d}{dy}\left\{\left[1+(y+\frac{1}{2})^2\right]^{-\frac{5}{2}}\left(y+\frac{1}{2}\right)+\left[1+(y-\frac{1}{2})^2\right]^{-\frac{5}{2}}\left(y-\frac{1}{2}\right)\right\}$ End your answer by showing that, $\frac{d^2B_x}{dy^2} = -3\frac{\mu_0I}{2}Na^{-1}\left\{\left[1+(y+\frac{1}{2})^2\right]^{-\frac{5}{2}}\left[1-5(y+\frac{1}{2})^2\right]\left[1+(y+\frac{1}{2})^2\right]^{-1}\right\}$ $\qquad+\left\{\left[1+(y-\frac{1}{2})^2\right]^{-\frac{5}{2}}\left[1-5(y-\frac{1}{2})^2\right]\left[1+(y-\frac{1}{2})^2\right]^{-1}\right\}\}$. 7. Evaluate your expression for $\frac{d^2B_x}{dy^2}$ at the center of convergence. Simplify. (After a few lines, you should get zero.) The result of the previous question is that the third term of the Taylor series is also zero. We get, $B_x = \frac{8\mu_0I}{a5\sqrt{5}}N + O[y]^3$. In the approximation above, you might need to remind yourself about which quantities are constants in this experiment and which are variable. Remember that the $O[y]^3$ reminds us that an exact expression for $B_x$ does involve $y^3$. However, there are no terms in $B_x$ that just involve either $y^1$ or $y^2$. Sometimes, the x-component of the field is written, $B_x \approx \frac{8\mu_0I}{a5\sqrt{5}}N$. (The "$\approx$" means "is approximately equal to".)

          $\frac{d^2B_x}{dy^2} = -3\frac{\mu_0I}{2}Na^{-1}\frac{d}{dy}\left\{\left[1+(y+\frac{1}{2})^2\right]^{-\frac{5}{2}}\left(y+\frac{1}{2}\right)+\left[1+(y-\frac{1}{2})^2\right]^{-\frac{5}{2}}\left(y-\frac{1}{2}\right)\right\}$
End your answer by showing that,
$\frac{d^2B_x}{dy^2} = -3\frac{\mu_0I}{2}Na^{-1}\left\{\left[1+(y+\frac{1}{2})^2\right]^{-\frac{5}{2}}\left[1-5(y+\frac{1}{2})^2\right]\left[1+(y+\frac{1}{2})^2\right]^{-1}\right\}$
$\qquad+\left\{\left[1+(y-\frac{1}{2})^2\right]^{-\frac{5}{2}}\left[1-5(y-\frac{1}{2})^2\right]\left[1+(y-\frac{1}{2})^2\right]^{-1}\right\}\}$.
7. Evaluate your expression for $\frac{d^2B_x}{dy^2}$ at the center of convergence. Simplify. (After a few lines, you should get zero.)
The result of the previous question is that the third term of the Taylor series is also zero. We get,
$B_x = \frac{8\mu_0I}{a5\sqrt{5}}N + O[y]^3$.
In the approximation above, you might need to remind yourself about which quantities are constants in this experiment and which are variable. Remember that the $O[y]^3$ reminds us that an exact expression for $B_x$ does involve $y^3$. However, there are no terms in $B_x$ that just involve either $y^1$ or $y^2$. Sometimes, the x-component of the field is written,
$B_x \approx \frac{8\mu_0I}{a5\sqrt{5}}N$.
(The "$\approx$" means "is approximately equal to".)

$(d^2Bx)/(dy^2) = -3(μ0I)/(2)Na^-1(d)/(dy){[1+(y+(1)/(2))^2]^-(5)/(2)(y+(1)/(2))+[1+(y-(1)/(2))^2]^-(5)/(2)(y-(1)/(2))} End your answer by showing that, (d^2Bx)/(dy^2) = -3(μ0I)/(2)Na^-1{[1+(y+(1)/(2))^2]^-(5)/(2)[1-5(y+(1)/(2))^2][1+(y+(1)/(2))^2]^-1} +{[1+(y-(1)/(2))^2]^-(5)/(2)[1-5(y-(1)/(2))^2][1+(y-(1)/(2))^2]^-1}}. 7. Evaluate your expression for (d^2Bx)/(dy^2) at the center of convergence. Simplify. (After a few lines, you should get zero.) The result of the previous question is that the third term of the Taylor series is also zero. We get, Bx = (8μ0I)/(a5√(5))N + O[y]^3. In the approximation above, you might need to remind yourself about which quantities are constants in this experiment and which are variable. Remember that the O[y]^3 reminds us that an exact expression for Bx does involve y^3. However, there are no terms in Bx that just involve either y^1 or y^2. Sometimes, the x-component of the field is written, Bx ≈(8μ0I)/(a5√(5))N. (The "≈" means "is approximately equal to".)$

Added by Sharon C.

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