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• Notes

$\begin{array}{l}{\text { As a new electrical technician, you are designing a large }} \\ {\text { solenoid to produce a uniform } 0.150-\text { T magnetic field near the cen- }} \\ {\text { ter of the solenoid. You have enough wire for } 4000 \text { circular turns. }} \\ {\text { This solenoid must be } 1.40 \mathrm{m} \text { long and } 2.80 \mathrm{cm} \text { in diameter. What }} \\ {\text { current will you need to produce the necessary field? }}\end{array}$

$B=M_{0} n I=M_{0} \frac{N}{L}I$

$\rightarrow I = \frac{B L}{M_{0} N}$$=\frac{0.15 * 1.4}{4 \pi*10^{-7} * 4000}=41.8 \mathrm{A}$

$\begin{array}{l}{\text { A } 15.0 \text { -cm-long solenoid with radius } 0.750 \mathrm{cm} \text { is closely }} \\ {\text { wound with } 600 \text { turns of wire. The current in the windings is } 8.00 \mathrm{A} \text { . }} \\ {\text { Compute the magnetic field at a point near the center of the }} \\ {\text { solenoid. }}\end{array}$

$B=M_{0} n I=\frac{M_{0} N I}{L}$

$B=\frac{M_{0}(600)(8)}{0.15}$

$=0.0402 \mathrm{T}$

$\begin{array}{l}{\text { A magnetic field of } 37.2 \text { T has been achieved at the MIT }} \\ {\text { Francis Bitter National Magnetic Laboratory. Find the current }} \\ {\text { needed to achieve such a field (a) } 2.00 \mathrm{cm} \text { from a long, straight }} \\ {\text { wire; (b) at the center of a circular coil of radius } 42.0 \mathrm{cm} \text { that has }} \\ {100 \text { turns; }(\mathrm{c}) \text { near the center of a solenoid with radius } 2.40 \mathrm{cm},} \\ {\text { length } 32.0 \mathrm{cm}, \text { and } 40,000 \text { turns. }}\end{array}$

(a) $B=\frac{M_{0}I}{2 \pi v} \rightarrow I=\frac{2 \pi v B}{M_{0}}$

$=\frac{2 \pi\left(2*10^{-2}\right)(37 \cdot 2)}{4 \pi * 10^{-7}}=3.72 * 10^{6} \mathrm{A}$

(b) $B=\frac{N M_{0} I}{2 R} \rightarrow I=\frac{2 R B}{N{M_{0}}}$

$=\frac{2(0.21)(37.2)}{100*4 \pi*10^{-7}}$$=1.24 * 10^{5} \mathrm{A}$

(c) $B=M_0 \frac{NI}{L}$$\longrightarrow \quad I=\frac{BL}{M_{0} N}$

$=\frac{37.2 * 0.32}{4 \pi * 10^{-7}+40000}=237A$