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www.GOALias.blogspot.comPhysics(iii) The electrostatic field is along the displacement vector joining thesource and the field point. The magnetic field is perpendicular to theplane containing the displacement vector r and the current elementI dl.(iv) There is an angle dependence in the Biot-Savart law which is notpresent in the electrostatic case. In Fig. 4.9, the magnetic field at anypoint in the direction of dl (the dashed line) is zero. Along this line,θ = 0, sin θ = 0 and from Eq. [4.11(a)], |dB| = 0.There is an interesting relation between ε 0, the permittivity of freespace; µ 0, the permeability of free space; and c, the speed of light invacuum:⎛ µ0 ⎞εµ0 0= ( 4πε0) ⎜⎝⎟4 ⎠19×10⎛ ⎞ −7=9 ( 10 )π⎜⎝⎟⎠1 1= =8 2 2(3 × 10 ) cWe will discuss this connection further in Chapter 8 on theelectromagnetic waves. Since the speed of light in vacuum is constant,the product µ 0ε 0is fixed in magnitude. Choosing the value of either ε 0orµ 0, fixes the value of the other. In SI units, µ 0is fixed to be equal to4π × 10 –7 in magnitude.Example 4.5 An element∆ l = ∆xˆi is placed at the origin and carriesa large current I = 10 A (Fig. 4.10). What is the magnetic field on they-axis at a distance of 0.5 m. ∆x = 1 cm.144EXAMPLE 4.5FIGURE 4.10Solutionµ d sin θB = [using Eq. (4.11)]π r0 I l|d | 42−2−7Tmdl=∆ x = 10 m , I = 10 A, r = 0.5 m = y, µ0/4π=10Aθ = 90° ; sin θ = 1−7 −210 × 10 × 10dB=−2= 4 × 10 –8 T25 × 10The direction of the field is in the +z-direction. This is so since,dl× =∆x ˆi× yˆj= y∆xˆi×j ˆ= y∆xkˆr ( )We remind you of the following cyclic property of cross-products,ˆi× ˆj = kˆ; ˆj× kˆ = ˆi;kˆ × ˆi = ˆjNote that the field is small in magnitude.