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Magnetic field of a circular loop
Appears in
Concepts tested here
- Current due to rotating charge 2
- Field of semicircular arcs
All Questions
2011 AIPMT-MAINS 1 question
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Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring isA rotating charged ring is equivalent to a circular current loop.
The charge q passes any point f times per second, so the current is $I = \frac{q}{T} = qf$
Magnetic field at the centre of a circular loop of radius R: $B = \frac{\mu_0I}{2R}$
$B = \frac{\mu_0qf}{2R}$
2010 AIPMT-MAINS 1 question
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A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i, the resultant magnetic field due to the two semicircular parts at their common centre isMagnetic field at the centre of a full circular loop is $\frac{\mu_0i}{2R}$; a semicircle gives half of this: $B_1 = B_2 = \frac{\mu_0i}{4R}$
The field of each semicircle is perpendicular to its own plane. The semicircle in the x-y plane gives a field along the z-axis, and the one in the x-z plane gives a field along the y-axis.
The two fields are therefore perpendicular to each other and equal in magnitude.
Resultant: $B = \sqrt{B_1^2 + B_2^2} = \sqrt{2}\times\frac{\mu_0i}{4R}$
$B = \frac{\mu_0i}{2\sqrt{2}R}$
2010 AIPMT-PRE 1 question
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A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of f revolutions/s. The value of magnetic induction in Wb/$m^2$ at the centre of the ring isA rotating charged ring is equivalent to a circular current.
The charge q crosses any point f times each second, so the current is $I = qf$
Field at the centre of a circular loop of radius R: $B = \frac{\mu_0I}{2R}$
$B = \frac{\mu_0qf}{2R}$