Question

A nonconducting ring of radius 10.0 cm is uniformly charged with a total positive charge 10.0μC. The ring rotates at a constant angular speed 20.0 rad/ s about an axis through its center, perpendicular to the plane of the ring. What is the magnitude of the magnetic field on the axis of the ring 5.00 cm from its center?

Answer 1

The magnitude of the magnetic field on the axis of the ring 5 cm from its center is 143 pT.

The radius of the nonconducting ring is R = 10 cm.

The ring is uniformly charged q = 10 μC.

The angular speed of the ring, ω = 20 rad/s

The ring is x = 5 cm from the center of the ring.

Now,

R = 10 cm = 0.1 m

q = 10.0 μC = 10 × 10⁻⁶ C

x = 5 cm = 0.05 m

The magnetic field on the axis of a current loop is given as:

B = [ μ₀ IR² ] / [4π(x² + R²)^{3/2} ]

Now, I = q / [2π/ω]

So, the magnitude of the magnetic field which is directed away from the center is:

B =  [ μ₀ ωqR² ] / [4π(x² + R²)^{3/2} ]

B = [ μ₀ (200) (10 × 10⁻⁶) (0.1)² ] / [4π((0.05)² + (0.1)²)^{3/2} ]

B = 1.43 × 10⁻¹⁰ T

B = 143 pT

Learn more about the magnetic field here:

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