The magnitude of the magnetic field at the centre of a circular current loop of radius R carrying current I is:
Aμ0 I/(2R)
Bμ0 I/(2πR)
Cμ0 I R/2
Dμ0 I²/(2R)
Answer & Solution
Correct answer: A. μ0 I/(2R)
B = μ0 I/(2R) at the centre.
Related questions
Two straight long conductors $AOB$ and $COD$ are perpendicular to each other and carry curA cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of centrAn infinitely long conductor $PQR$ is bent to form a right angle as shown. A current $I$ fFor a positively charged particle moving in a $x - y$ plane initially along the $x$-axis, A small square loop of wire of side $l$ is placed inside a large square loop of wire of siA wire of length $1\,\mathrm{m}$ is moving at a speed of $2\,\mathrm{ms}^{-1}$ perpendiculA uniform but time-varying magnetic field $B(t)$ exists in a circular region of radius $a$The unit ratio μ0 ε0 has the same dimensions as: