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  4. Chapter 21 Maxwell’s equations; electromagnetic radiation
  5. Chapter 21 problems
  6. Answer to Chapter 21 problems

Answer to Chapter 21 problems

21.1.1   (a)  6.2 mA    (b)  2.5 nT

21.1.2   0.4 μV

21.1.3   6.25 × 10–11 T;    2.5 × 10–10  T;    1.0 × 10–10  T

21.2.1   2.2 × 10–19

21.2.2   0.5 nT

21.2.3   2.0 × 108  m s–1;    5.6 mA m–1

21.5.1   0.6

21.5.2   0.052;   horizontally

21.6.1   (a) 8.0 × 10–9 T     (b)  7.64 × 10–3 T;     6.15 W

21.6.2   0.498 km2;    both gravitational attraction and energy flux fall inversely with the square of r

21.6.3   <S> = \( {I^2 R \over 2 \pi rL} \);  the surface area of the wire = A = 2Ï€rL, so SA = I2R, that is, the rate at which electromagnetic energy flows into the wire is equal to the rate at which energy is dissipated by the wire

21.6.4  

21.6.5   (i) 0.637 W m–2     (ii)  2.12 Ã— 10–9 Pa      (iii)  7.31 × 10–8 T      (iv)  21.9 V m–1

21.6.6   9.21 h

21.6.7   (i) 0.105 kg   (ii)  1.58 x 104 m s–1   (iii)  2.50 × 1011 m     (iv) 2.54 × 10–5

21.6.8   980 V m–1;    3.3 mT  

21.6.9   730 V m–1;    1.9 × A m–1

21.6.10   7.9 mW m–2

21.6.11   1.7 nT;   830 kW

21.6.12   0.2 µm

21.6.13   6 × 107 V m–1;    1.6 × 105 A m–1;   17 kJ m–3;     34 mN

21.6.14   9 mm s–1

21.7.1   1.12 × 104 s–1;     2.75 × 10–4 V m–1

21.7.2   3.2 × 1016 

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