189 
1910-11.] Young’s Modulus under an Electric Current. 
^-component of magnetic force at O due to upward current idxdy 
_ 2 idxdy y 
r r 
. *. Total ic-component due to current through area xy is 
Put x = y tan <f>. 
.\dx — y sec 2 cf>d<j). 
\ x — iy j (log cosec 2 <£) sec 2 <f>d<fi 
— — iy j log sin 2 cfjd. (tan </>) 
= - fc’^tan </> log sin 2 </> 
The limits are $ — 0 and 0 = tan 1 — . 
y 
• i x , o • -i a; 
- ix log — + 2 iy tan — 
r 2 y 
= + | ix log ^ 1 + + 2 iy tan 1 
x 
y - 
Similarly, 
I, = - 
iy log ( 1 + — ) + 2 ix tan 
i V 
where I ?/ is positive outwards from O. 
To obtain the force at P due to the current in the strip, find the com- 
ponents along PD and PB. 
To get the components along PD — 
For rectangle (1) let PC be x and PA be y and use — I*. 
(2) „ 
PA 
„ X 
„ PD 
» y » 
„ +i,. 
„ 
(3) „ 
PD 
„ X 
„ PB 
» y » 
j> +i x - 
V 
5? 
(4) „ 
PB 
„ X 
„ PC 
« y ” 
„ -i. 
Add these, and we get the resultant along PD. Let us call it F r 
