Answer :

Given:


Number of turns, N = 20


Radius of coil, r = 10cm


Magnetic field strength = 0.10 T


Current in the coil = 5 A



a) The torque on the coil is zero because the magnetic field is uniform along the coil.


b) The total force on the coil is Zero due to uniform magnetic field.


c) Let average force on each electron in the coil due to magnetic field = F


Cross-sectional area of the wire, A = 10-5 m2


Free electron density or number of free electron per metre cube of copper, D = 1029


Charge in each electron = 1.6 × 1019 C


We know that,


…(1)


Where,


F = force due to magnetic field


B = Magnetic field strength


e = charge on electron


Vd = drift velocity of electron


Vd = I/ (NeA) …(2)


Where, I = current through the coil


N = Density of free electron


A = cross sectional area.


Putting values in equation (2)




Vd = 3.125 × 1043 ms-1


F = B × e × Vd …(3)


By putting the values of B, e and Vd in equation 3, we get,



F = 5 × 10-25 N


Hence, average force on each electron is 5 × 10-25 N.


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