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www.GOALias.blogspot.comPhysicsExample 4.9 A solenoid of length 0.5 m has a radius of 1 cm and ismade up of 500 turns. It carries a current of 5 A. What is themagnitude of the magnetic field inside the solenoid?Solution The number of turns per unit length is,EXAMPLE 4.9500n = = 1000 turns/m0.5The length l = 0.5 m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a.Hence, we can use the long solenoid formula, namely, Eq. (4.20)B = µ 0n I= 4π × 10 –7 × 10 3 × 5= 6.28 × 10 –3 T4.9 FORCE BETWEEN TWO PARALLEL CURRENTS,THE AMPEREWe have learnt that there exists a magnetic field due to a conductorcarrying a current which obeys the Biot-Savart law. Further, we havelearnt that an external magnetic field will exert a force ona current-carrying conductor. This follows from theLorentz force formula. Thus, it is logical to expect thattwo current-carrying conductors placed near each otherwill exert (magnetic) forces on each other. In the period1820-25, Ampere studied the nature of this magneticforce and its dependence on the magnitude of the current,on the shape and size of the conductors as well as thedistances between the conductors. In this section, weshall take the simple example of two parallel currentcarryingconductors, which will perhaps help us toappreciate Ampere’s painstaking work.Figure 4.20 shows two long parallel conductors aFIGURE 4.20 Two long straightand b separated by a distance d and carrying (parallel)parallel conductors carrying steadycurrents I currents I aand I b, respectively. The conductor ‘a’aand I band separated by adistance d. B ais the magnetic field set produces, the same magnetic field B aat all points alongup by conductor ‘a’ at conductor ‘b’. the conductor ‘b’. The right-hand rule tells us that thedirection of this field is downwards (when the conductorsare placed horizontally). Its magnitude is given by Eq. [4.19(a)] or fromAmpere’s circuital law,Ba=µ0Ia2 π d154The conductor ‘b’ carrying a current I bwill experience a sidewaysforce due to the field B a. The direction of this force is towards theconductor ‘a’ (Verify this). We label this force as F ba, the force on asegment L of ‘b’ due to ‘a’. The magnitude of this force is given byEq. (4.4),