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Calculate the magnetic field inside a solenold ,when (a) The length of the solenoid becomes twice and fixed number of turns (b) both the length of the solenoid and number of turns are double (c ) the number of turns becomes twice for the fixed length of the solenoid compare the results, |
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Answer» Solution :The MAGNETIC field of a solenoid (inside ) is `B_(L.N) = mu_(0) (NI)/(L)` (a) length of the solenoid becomes TWICE and FIXED number of turns L `rarr`2L (length becomes twice ) N `rarr` N(number of turns are fixed ) The magnetic field is `B_(2 L,N) = mu_(0) (NI)/(2L) = (1)/(2) B_(L,N)` (b) both the length of the solenoid and number of turns are double L `rarr` 2L (length becomes twice ) N `rarr` 2N ( number of turns becomes twice ) The magnetic field is `B_(2L,2N) = mu_(0) (2NI)/(2L) = B_(L,N)` ( c) the number of turns becomes twice but for the fixed lenght of the solenoid . L `rarr` L (length is fixed ) N `rarr` 2N ( number of turns becomes twices ) the magnetic field is `B_(L,2N) = mu_(0) (2NI)/(L) = 2B_(L,N)` From the above results , `B_(2L,2N) gt B_(2L,2N) gt B_(2L,N)` THUS, strength of the magnetic field is increased when we PACK more loops into the same length for a given current. |
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