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(a) Find magnetic field inside a long solenoid, carrying current (1//pi) A and having number of turns per unit length 10. Also calculate magnetic field at ends. (b) A long solenoid is fabricated by closely winding a wire of diameter 10 mm over a cylinderical non-magnetic frame so that the successive turns nearly touch each other. Find magnetic field at the centre and ends of solenoid if it carriers a current (4//pi) A? (c) A single-layer coil (solenoid) has length l and cross-section radius R. The number of turns per unit length is equal to n. Find the magnetic induction at the centre of the coil when a current i flows through it. (d) A solenoid of length 0.4 m anddiameter 0.6 m consists of a single layer of 1000 turns of the fine wire carrying a current of 5 mA. Calculate the magnetic field on the axis at the middle and at the ends of the solenoid. |
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Answer» Solution :(a) `i=1/piA, n=10` `B_(centre)=mu_(0)ni=(4pixx10^(-7))(10)(1/pi)=4xx10^(-6) T` `B_(end)=1/2 mu_(0) ni=2xx10^(-6) T` (b)  Number of TURNS in `10 mm` length `=1` Number of turns in `1m` length `n=100` `i=4/(pi) A` `B_(centre)=mu_(0)ni=(4pixx10^(-7))(100)(4/pi)` `=16xx10^(-5)T` `B_(end)=1/2 mu_(0) ni =8xx10^(-5) T` (C)  `cos theta_(1)=cos theta_(2)=(l//2)/(sqrt(R^(2)+(l//2)^(2)))=l/(sqrt(l^(2)+4R^(2)))` `B_(C)=1/2mu_(0)ni(cos theta_(1)+cos theta_(2))` `=1/2 mu_(0)nixx(2l)/(sqrt(l^(2)+4R^(2)))` `=(mu_(0)nil)/(sqrt(l^(2)+4R^(2)))=(mu_(0)ni)/(sqrt(l+(2R//l)^(2)))` If `l to oo, B_(C)=mu_(0)ni` (d) `i=5mA=5xx10^(-3) A, m=N//l=1000/0.4=2500`  `cos theta_(1)=0.2/(sqrt((0.2)^(2)+(0.3)^(2)))=0.2/(sqrt(0.13))=2/(sqrt(13))` `B_(C)=1/2mu_(0)ni(cos theta_(1)+cos theta_(2))` `=1/2(4pixx10^(-7))(2500)(5xx10^(-3))xx2xx2/(sqrt(13))` `=(pixx10^(-5))/(sqrt(13)) T` At ENDS  `cos THETA=0.4/0.5=4/5=0.8` `B_(A)=1/2mu_(0)ni(cos 90^(@)+cos theta)` `=1/2(4pixx10^(-7))(2500)(5xx10^(-3))XX(0.8)` `=2pixx10^(-6) T` |
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