TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Radius of the loop, r=1cm=0.01m
Total charge on the loop, Q=1×10−6C
∴‌ Charge per unit length, ‌λ=‌

Q

2πr

=‌

10−6

2π×0.01

‌‌=‌

10−4

2π

C∕m
r′=0.01%‌ of the length ‌=‌

0.01

100

×2πr
⇒‌‌l′=‌

2π×0.01×0.01

100

‌‌=‌‌2π×10−6m
For the whole circular loop, the electric field at the centre is zero due to symmetry. But when a certain portion of the loop is cut off, then there will exist an electric field at the centre due to that cutting part. Which is equal to the electric field due to remaining part of wire as per charge conservation.
∴‌ Charge of the part ‌l′,Q′‌‌=1λ=‌

2π×10−6×10−4

2π

‌‌=10−10C
∴‌ Electric field at centre, ‌E‌‌=‌

Q′

4πε0r2

E‌‌=‌

10−10×9×109

(0.01)2

‌‌=9×103N∕C