Q. 444.4( 11 Votes )

# A circular wire-l

Answer :

**Given:**Radius of the circular loop : a

Total charge on the wire : Q

Length of the cut off wire: dL

**Formula used:**

Electric field is given as:

Here, k is a constant and k= =9× 10

^{9}Nm

^{2}C

^{-2}. q is the point charge and r is the distance between the charge and the point of influence.

Here r=a=radius of the loop.

**We know that, electric field at the center of the uniformly charged circular wire**

Which means that sum of electric field due to cut off wire and remaining wire is zero.

We know that,

Where λ is the linear charge density. Q is the total charge and L is the length of the wire.

Charge on the element dL be dq:

Here L = 2πa=circumference. Here, a is the radius of the loop and L is the Length of the loop.

Thus, electric field due to dL(cutoff wire) at the center is:

Since

Thus, magnitude of electric field at the center of the circular wire due to remaining wire is E= but in opposite direction to that of the field due to cut off wire.

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