Solutions 060519

Question 1

The two fragments can be treated as point charges for purposes of calculating their potential energy. Using energy conservation, the potential energy is all converted to kinetic energy as the two fragments separate to a large distance.

$E_{\textrm {initial}}=E_{\textrm {final}}\rightarrow U_{\textrm {initial}}=K_{\textrm {final}}={\frac {1}{4\pi \epsilon _{0}}}{\frac {q_{1}q_{2}}{r}}V$

$=8.99\times 10^{9}{\textrm {N}}{\textrm {m}}^{2}/{\textrm {C}}^{2}{\frac {(38)(54)(1.60\times 10^{-19}{\textrm {C}})^{2}}{(5.5\times 10^{-15}{\textrm {m}})+(6.2\times 10^{-15}{\textrm {m}})}}{\frac {1{\textrm {eV}}}{1.60\times 10^{-19}{\textrm {J}}}}=250\times 10^{6}{\textrm {eV}}=250{\textrm {MeV}}$

This is about %25 greater than the observed kinetic energy of 200 MeV

Question 2

(a)

Set $x=0$ at the midpoint on the axis between the two loops. Since the loops are a distance $R$ apart, the center of one loop will be at $x=-{\frac {1}{2}}R$ and the center of the other at $x=+{\frac {1}{2}}R$ . The currents in the loops flow in opposite directions, so by the right-hand-rule the magnetic fields from the two wires will subtract from each other.

for the single coil:

$dB={\frac {\mu _{0}Idl}{4\pi r^{2}}}$

$B=B_{\parallel }=\int dB\cos \phi =\int dB{\frac {R}{r}}=\int dB{\frac {R}{(R^{2}+x^{2})^{\frac {1}{2}}}}={\frac {\mu _{0}I}{4\pi r^{2}}}{\frac {R}{(R^{2}+x^{2})^{\frac {1}{2}}}}\int dl={\frac {\mu _{0}IR^{2}}{2(R^{2}+x^{2})^{\frac {3}{2}}}}$

In the anti-helmholtz setup:

$B(x)={\frac {\mu _{0}NIR^{2}}{2\left[R^{2}+\left({\frac {1}{2}}R-x\right)^{2}\right]^{\frac {3}{2}}}}-{\frac {\mu _{0}NIR^{2}}{2\left[R^{2}+\left({\frac {1}{2}}R+x\right)^{2}\right]^{\frac {3}{2}}}}$

$B(x)={\frac {\mu _{0}NI}{R\left[4+\left(1-2x/R\right)^{2}\right]^{\frac {3}{2}}}}-{\frac {\mu _{0}NI}{R\left[4+\left(1+2x/R\right)^{2}\right]^{\frac {3}{2}}}}$

$={\frac {4\mu _{0}NI}{R}}\left\{\left[4+\left(1-{\frac {2x}{R}}\right)^{2}\right]^{-{\frac {3}{2}}}-\left[4+\left(1+{\frac {2x}{R}}\right)^{2}\right]^{-{\frac {3}{2}}}\right\}$

(b)

For small values of $x$, $x^{2}\approx 0$ so $\left(1\pm {\frac {2x}{R}}\right)^{2}\approx =1\pm {\frac {4x}{R}}$

$B(x)={\frac {4\mu _{0}NI}{R}}\left\{\left[4+\left(1-{\frac {4x}{R}}\right)\right]^{-{\frac {3}{2}}}-\left[4+\left(1+{\frac {4x}{R}}\right)\right]^{-{\frac {3}{2}}}\right\}={\frac {4\mu _{0}NI}{5R{\sqrt {5}}}}\left\{\left[1-{\frac {4x}{5R}}\right]^{-{\frac {3}{2}}}-\left[1+{\frac {4x}{5R}}\right]^{-{\frac {3}{2}}}\right\}$

Again we can use the expansion for small deviations $\left(1\pm {\frac {4x}{5R}}\right)^{-{\frac {3}{2}}}\approx \left(1\mp {\frac {6x}{5R}}\right)$

$B(x)={\frac {4\mu _{0}NI}{5R{\sqrt {5}}}}\left[\left(1+{\frac {6x}{5R}}\right)-\left(1-{\frac {6x}{5R}}\right)\right]={\frac {48\mu _{0}NIx}{25R^{2}{\sqrt {5}}}}$

The linear coefficient is :

$B(x)=Cx$

$C={\frac {48\mu _{0}NI}{25R^{2}{\sqrt {5}}}}$

(c)

Set C equal to 0.15 T/m and solve for the current.

$I={\frac {25CR^{2}{\sqrt {5}}}{48\mu _{0}N}}={\frac {25(0.15{\textrm {T/m}})(0.04{\textrm {m}})^{2}{\sqrt {5}}}{48(4\pi \times 10^{-7}{\textrm {Tm/A}})(150)}}=1.5{\textrm {A}}$