An electron is shot into one end of a solenoid . As it enters the uniform magnet

Question

An electron is shot into one end of a solenoid . As it enters the uniform magnetic field within the solenoid, its speed is 800 m/s and its velocity vector makes an angel of 30 degree with the central axis of the solenoid . The solenoid carries 4.0A and has 8000 turns along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid's opposite end ?(In a real solenoid, where the field is not uniform at the two ends the number of revolutions would be slightly less than the answer here.) SSM ILW WWW A long solenoid with 10.0 turns/cm and a radius of 7.00 cm carries a current of 20.0 mA . A current of 6.00 A exists in a straight conductor located along the central axis of the solenoid. At what redial distance from the axis will the direction of the resulting magnetic field be at 45 degree to the axial direction ? What is the magnitude of the magnetic field there? Sec.29-6 A current - carrying coil as a Magnetic Dipole Figure 29-71 shows an arrangement known as a Helmholtz coil . It consist of two circular coaxial coils, each of 200 turns and radius

Explanation / Answer

a)

B1 = u0 N i/L = 4*3.1416e-7*(10/0.01) * 20e-3 = 0.0000251328 T

B2 = u0 i2/2pir

0.0000251328 = 2e-7*6/r

r = 0.001326 = 0.00133 m

b)

B = sqrt(2) * 0.0000251328 = 3.55E-5 T

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