HW08sol - Classical Electrodynamics I PHY 505 Fall 2010...

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1 C:\User\Teaching\505-506\F10\HW08sol.doc PHY 505 Classical Electrodynamics I Fall 2010 Homework 8 with solutions Problem 8.1 (to be graded of 10 points). Calculate the magnetic field distribution along the axis of a straight solenoid (Fig. 5.6a of the lecture notes) with a finite length l, and a round cross-section of radius R . Assume that the solenoid has many wire turns ( N >> 1) which are uniformly distributed along its length. Solution : Because of the large number of wire turns, and hence the almost circular shape of each turn, the solenoid field may be presented as a sum of field of N circular current loops. The field induced by a single loop has been calculated in class - see Eq. (5.23) of the lecture notes. Generalizing it to the case of arbitrary position ( z’ ) of the loop at the z -axis, we get  2 / 3 2 2 2 0 1 ) ' ( 2 z z R R I B . Now we can find induced by the solenoid as a whole as   2 / 2 / 2 / 3 2 2 2 0 2 / 2 / 2 / 3 2 2 2 0 ) / ( 2 ) ( 2 N N j N N j j N N jl z R R I z z R R I B , where index j numbers wire turns (starting from the middle of the solenoid). For N >> 1 this sum is equivalent to integral 1   . ) 2 / ( 2 / ) 2 / ( 2 / 2 ) ' ( ' 2 ' ) ' ( 2 ) / ( 2 2 2 2 2 0 2 / ' 2 / ' 2 2 0 2 / 2 / 2 / 3 2 2 2 0 2 / 2 / 2 / 3 2 2 2 0 l z R l z l z R l z l N I z z R z z l N I dz z z R R l N I dj N jl z R R I B l z l z l l N N N Well inside a long solenoid ( l >> z , R ), this expression reduces to Eq. (5.40). On the other hand, at z >> R , l , this result gives 1 2 3 0 3 2 0 , 2 4 2 Nm INR m z m z INR B N N N , i.e. the dipole field, with the dipole moment

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HW08sol - Classical Electrodynamics I PHY 505 Fall 2010...

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