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A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ? |
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Answer» The number of turns per unit length is `n=(500)/(0.5)=1000` turns/ m The length l=0.5m and radius r=0.01 m Thus, `l//a=50" i.e.,"l gt gt a`. Hence, we can use the long solenoid formula, namely, Eq. `(B= mu_(0) nI) B=mu_(0)nI` `=4pi xx 10^(-7)xx10^(3)xx5=6.28xx10^(-3) T` |
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