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I Have an open challenge for science students..!!!A solenoid has 2 x 10^4 turns/meter and hasdiameter 10 cm. An electron beam having K.E. 100kev passes without touching walls of solenoid thenfind current in the solenoid. Electron beam makeangle 30° with axis of solenoid,(1) 0.4 A(2) 0.8 A(3) 1.2 A(4) 1.6 Aremember its a challenge! |
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Answer» Answer: We are given: Number of turns of solenoid, N = 100 Length of the solenoid, L = 20 cm = 0.2 m diameter of the solenoid, d = 4 cm = 0.04 m Current through the solenoid, I = 2 A Cross sectional area of the solenoid, A =πd24 =π×(0.04)24 =1.26×10−3 m2A =πd24 =π×(0.04)24 =1.26×10−3 m2 (a) Finding the magnetic field (B) in the core of the solenoid B =μo×NL×IB =μo×NL×I B =4π×10−7×1000.2×2B =4π×10−7×1000.2×2 B =1.26×10−3 TB =1.26×10−3 T (b) Finding the inductance (L) of the solenoid L =μo×N2L×AL =μo×N2L×A L =4π×10−7×10020.2×1.26×10−3L =4π×10−7×10020.2×1.26×10−3 L =7.92×10−5 HL =7.92×10−5 H (C) Magnetic ENERGY stored (E) in the solenoid E =12×L×I2E =12×L×I2 E =12×7.92×10−5×22E =12×7.92×10−5×22 E =1.58×10−4 JE =1.58×10−4 J |
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