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$\begin{array}{l}{\text { A closely wound, circular coil with radius } 2.40 \mathrm{cm} \text { has }} \\ {800 \text { turns. (a) What must the current in the coil be if the magnetic }} \\ {\text { field at the center of the coil is } 0.0580 \mathrm{T} \text { ? (b) At what distance } x} \\ {\text { from the center of the coil, on the axis of the coil, is the magnetic }} \\ {\text { field half its value at the center? }}\end{array}$

$X=0, \quad B_ x=\frac{M_{0} N I}{2 a}$

(a) $\rightarrow I=\frac{2 a B_{x}}{M_{0} N}=\frac{2(0.024)(0.058)}{4 \pi * 10^{-7} * 800}$

$=2.77 \mathrm{A}$

(b) $B_{x}=\frac{M_{0} N I a^{2}}{2\left(x^{2}+a^{2}\right)^{3/2}}=$$\left(\frac{M_{0} NI }{2 a}\right)\left(\frac{a^{3}}{\left(x^{2}+a^{2}\right)^{3 / 2}}\right)$$=B_{c}\left(\frac{a^{3}}{\left(x^{2}+a^{2}\right)^{3/2}}\right)$

$B_ x=\frac{1}{2} B_c$$\quad \rightarrow \frac{1}{2} B_ c=$$B_ c\left(\frac{a^{3}}{\left(x^{2}+a^{2}\right)^{3/2} }\right)$

$∴ \quad \frac{1}{2}=\frac{a^{3}}{\left(x^{2}+a^{2}\right)^{3 / 2}} \rightarrow\left(x^{2}+a^{2}\right)^{3}=4 a^{6} \rightarrow x=0.084 m$

$\begin{array}{l}{\text { A closely wound, circular coil with a diameter of } 4.00 \mathrm{cm}} \\ {\text { has } 600 \text { turns and carries a current of } 0.500 \mathrm{A} \text { . What is the magni- }} \\ {\text { tude of the magnetic field (a) at the center of the coil and (b) at a }} \\ {\text { point on the axis of the coil } 8.00 \mathrm{cm} \text { from its center? }}\end{array}$

(a) $B_{\text { canter }}=\frac{M_{0} N I}{2 a}$

$=\frac{4 \pi * 10^{-7} * 600*0.5}{2(0.02)}=9.42*10^{-3} \mathrm{T}$

(b) $B(x)=$$=\frac{M_{0} N I{a}^{2}}{2\left(x^{2}+a^{2}\right)^{\frac{3}{2}}}$

$=\frac{4 \pi*10^{-7} * 600 * 0.5*0.02^{2}}{2\left(0.08^{2}+0.02^{2}\right)^{3 / 2}}=1.34*10^{-4} T$