At position P1
: due to I1,calculate B1 = k.I1/r = (2 x 10-7 x 12)/4 x 10-2 = 6.0 x 10-5 T down.
: similarly due to I2 = B2 = 1.0 x 10-4 T up.
Hence the total field at P1 added vectorially is 4.0 x 10-5 T up.
At position P2
: due to I1 calculate B2 = (2 x 10-7 x 12)/4 x 10-2 = 6.0 x 10-5 T down.
: due to I2 = 1.0 x 10-4 T down.
Hence the total field at P2 added vectorially is 1.6 x 10-4 T down.
The plot of Force verses current is shown below as an Excel chart, with an included trend line of best fit.
(a) The weight of the coil will be the force value when coil current is zero ie a value of 3.0 N.
(b) the magnetic field strength (B) can be found from the slope of the curve as F = B I L (N) and hence slope = DF/DI = B L (N) if the field is a constant :-
substitution of the coil dimension and turns yields a value of B = 1.3 N.A-1.
(c) The current as determined from the right-hand-rule will be anticlockwise as seen from the magnet's south pole, because the force exerted by the current-carrying coil is downwards.
(d) From the graph above, with a force of zero, the current flowing in a clockwise direction must be 2.1 amps.
Given electrode-filament PD = 200V and dealing with electrons :-
(a) The magnetic field must be directed into the page.
(b) Kinetic energy Ek = q.V = 1.6 x 10-19 x 200 = 3.2 x 10-17 J = 200 eV.
(c) Equating energy 1/2.m v2 = q.V and using the radius of curvature formula,
then mass of the electron m = q.B2. r2/2.V
(d) Given the values B = 0.02T, and r = 2.5 mm and by substitution the electron mass calculates as 1.0 x 10-30 kg.
100 turn coil, B = 25 x 10-3 T and V = 120 V.
(a) Total coil resistance R = 100 x 0.018 W = 1.8 W.
Using Ohm's law I = V/R = 120/1.8 = 67A.
(b) Torque on the coil t = B.A.I.N where A = rectangular (LxB)area of coil.
By substitution t = 25 x 10-3 x 0.32 x 0.1 x 67 x 100 = 5.4 Nm.