446 Induction of Currents in Linear Circuits [CH. xiv 



A Iterna ting Curren t. 



514. Let us next suppose that the electromotive force in the circuit is 

 not produced by batteries, but by moving the circuit, or part of the circuit, 

 in a magnetic field. If N is the number of tubes of induction of the exter- 

 nal magnetic field which are enclosed by the circuit at any instant, the 

 equation is 



-j t (Li l + N) = Ri l ........................ (435). 



The simplest case arises when N is a simply-harmonic function of the 

 time, proportional let us say to cospt. We can simplify the problem by sup- 

 posing that N is of the form C (cospt + isiupt). The real part of N will 

 give rise to a real value of i lt and the imaginary part of N to an imaginary 

 value of v Thus if we take N = Ce ipt we shall obtain a value for i\ of which 

 the real part will be the true value required for i lf 



Assuming N = (7 (cos pt + isinpt) = Ce ipt , the equation becomes 



and clearly the solution will be proportional to e ipt . Thus the differential 



operator -y- will act only on a factor e ipt , and will accordingly be equivalent to 

 diii 



multiplication by ip. We may accordingly write the equation as 



a single algebraic equation of which the solution is 



. _-ipCe ipt 



Let the modulus and argument of this expression be denoted by p and %, 

 so that the value of the whole expression is p (cos ^ 4- i sin ^). The value of 

 p, the modulus, is equal ( 311) to the product of the moduli of the factors, so 

 that 



pC 



p vjF + zy' 



while the argument %, being equal ( 311) to the sum of the arguments of 

 the factors, is given by 



The solution required for ^ is the real term p cos %, so that 



Ij = p COS X 



.-!(& 



Jt 



