Question
Question: Charge q is uniformly spread on a thin radius R. The ring rotates about its axis with a uniform freq...
Charge q is uniformly spread on a thin radius R. The ring rotates about its axis with a uniform frequency fHz. The magnitude of magnetic induction at the center of the ring is
A. 2Rμ0qf
B. 2fRμ0q
C. 2πfRμ0q
D. 2πRμoqf
Solution
We solve this problem by using the concept of the magnetic field at the center of a current-carrying loop. Which are the applications of Biot- Savart’s law. If the ring is rotated in its axis then it forms a circular loop.
Complete step by step answer:
Consider a current flowing in the circular loop I = qf
(Because ring rotates about its axis with uniform frequency ⇒T1=f )
Now, the magnitude of induction at the centre of the ring is
B = 2RμoI=2Rμoqf .
(Here B is the magnetic field, R is the radius, I is the current and q is the charge.)
Also, μ0 is the absolute permeability and f is the frequency.
∴ Hence the correct option is (A).
Additional information:
Biot-savart’s law for the magnetic field obeys the inverse square law. That is the magnetic field due to the current element at a point is inversely proportional to the square of the distance between the current element and point of observation. Also, permeability is the capability of the substance to have magnetization in the presence of a magnetic field.
Note:
The direction of the magnetic field at the center of the current-carrying loop is perpendicular to the plane of the loop (ring) in the downward direction. If the direction of the magnetic field at the center of the current-carrying loop is perpendicular to the plane of the loop in the upward direction and the current passes anticlockwise.