Question #118253

Given a short solenoid, length 30cm,radius 15 cm ,500 turns,current 7 ampere. What is magnetic field at the end of the solenoid?


1
Expert's answer
2020-05-26T12:53:16-0400

Expression for the magnetic field at distance D from the edge of the solenoid can be given as


B=μNI2(D+L(D+L)2+R2DD2+R2)B=\frac{\mu NI}{2}\left(\frac{D+L}{\sqrt{(D+L)^2+R^2}}-\frac{D}{\sqrt{D^2+R^2}}\right)

Now at the end of the solenoid D=0, so expression becomes


B=μNI2(L(L)2+R2)B=\frac{\mu NI}{2}\left(\frac{L}{\sqrt{(L)^2+R^2}}\right)

B=(4π107)(500)(7)2(0.3(0.3)2+0.152)=0.002 T=2.0 mTB=\frac{(4\pi \cdot10^{-7})(500)(7)}{2}\left(\frac{0.3}{\sqrt{(0.3)^2+0.15^2}}\right)\\=0.002\ T=2.0\ mT


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