# Chapter 25 - Worked Solutions

### to questions from the OUP text *Senior Physics -
Concepts in Context* by Walding,
Rapkins and Rossiter

My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.

Solutions-Ch25-Finch.pdf

The rest are from Glenn Rossiter:

**Q14 :-**

At position P_{1}

: due to I_{1},calculate B_{1} = k.I_{1}/r = (2 x 10^{-7} x 12)/4 x 10^{-2} = 6.0 x 10^{-5} T down.

: similarly due to I_{2} = B_{2} = 1.0 x 10^{-4} T up.

Hence the total field at P_{1} added vectorially is 4.0 x 10^{-5} T up.

At position P_{2}

: due to I_{1} calculate B_{2} = (2 x 10^{-7} x 12)/4 x 10^{-2} = 6.0 x 10^{-5} T down.

: due to I_{2} = 1.0 x 10^{-4} T down.

Hence the total field at P_{2} added vectorially is 1.6 x 10^{-4} T down.

**Q32 :-**
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.

**Q33 :-**
Given electrode-filament PD = 200V and dealing with electrons :-

(a) The magnetic field must be directed into the page.

(b) Kinetic energy E_{k} = q.V = 1.6 x 10^{-19} x 200 = 3.2 x 10^{-17} J = 200 eV.

(c) Equating energy 1/2.m v^{2} = q.V and using the radius of curvature formula,

then mass of the electron m = q.B^{2}. r^{2}/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.

**Q35 :-**
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.

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