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    I was confused on this question in june 2014 unit 4 paper on aqa a physics.
    2 (b) (iii) With a current of 0.38 A, the average velocity of an electron in the wire is5.5 × 10^–6m s–1 and the average magnetic force on one electron is 1.4 × 10^–25N.Calculate the flux density B of the magnetic field.
    I was given that the wire is 95 mm long with 4.0x10^22 electron in the wire
    I don't understand if i used F=BQV i don't need to x 4.0x10^22 but for F=BIL i need to time 4.0x10^22 number of electron.
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    (Original post by John doe123543)
    Hi
    I was confused on this question in june 2014 unit 4 paper on aqa a physics.
    2 (b) (iii) With a current of 0.38 A, the average velocity of an electron in the wire is5.5 × 10^–6m s–1 and the average magnetic force on one electron is 1.4 × 10^–25N.Calculate the flux density B of the magnetic field.
    I was given that the wire is 95 mm long with 4.0x10^22 electron in the wire
    I don't understand if i used F=BQV i don't need to x 4.0x10^22 but for F=BIL i need to time 4.0x10^22 number of electron.
    In the first equation F is the force on a single electron. In the second, F is the force on the wire, hence you have to multiply the given force by the total number of electrons.
 
 
 
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