Question

A toroid with 500 turns of wire carries a current of (2\(\pi\)) Ampere. A metal ring inside the toroid provides the core and has susceptibility of 2x10^-5. If the magnetization 5x10^-2 A/m, then radius of the ring is:

• A. 50 cm
• B. 20\(\pi\) cm
• C. \(\frac{50}{\pi}\) cm
• D. 20 cm
• E. 60 cm

Answer 1

The Correct Option is D

Given:

• Number of turns in toroid, \( N = 500 \)
• Current through wire, \( I = 2\pi \, \text{A} \)
• Magnetic susceptibility, \( \chi = 2 \times 10^{-5} \)
• Magnetization, \( M = 5 \times 10^{-2} \, \text{A/m} \)

Step 1: Relate Magnetization to Magnetic Field

Magnetization (\( M \)) is given by:

\[ M = \chi H \]

where \( H \) is the magnetic field intensity.

Solving for \( H \):

\[ H = \frac{M}{\chi} = \frac{5 \times 10^{-2}}{2 \times 10^{-5}} = 2.5 \times 10^{3} \, \text{A/m} \]

Step 2: Express \( H \) for a Toroid

For a toroid, the magnetic field intensity is:

\[ H = \frac{NI}{2\pi r} \]

where \( r \) is the radius of the ring.

Step 3: Solve for Radius (\( r \))

Substitute the known values into the equation for \( H \):

\[ 2.5 \times 10^{3} = \frac{500 \times 2\pi}{2\pi r} \]

Simplify the equation:

\[ 2.5 \times 10^{3} = \frac{500}{r} \]

Solve for \( r \):

\[ r = \frac{500}{2.5 \times 10^{3}} = 0.2 \, \text{m} = 20 \, \text{cm} \]

Conclusion:

The radius of the ring is 20 cm.

Answer: \(\boxed{D}\)

Answer 2

Step 1: Recall the relationship between magnetization, susceptibility, and magnetic field.

The magnetization \( M \) of a material is related to its magnetic susceptibility \( \chi_m \) and the magnetic field \( H \) by the equation:

\[ M = \chi_m H, \]

where:

• \( M \) is the magnetization (in A/m),
• \( \chi_m \) is the magnetic susceptibility (dimensionless), and
• \( H \) is the magnetic field strength (in A/m).

 

We are given:

• \( M = 5 \times 10^{-2} \, \text{A/m}, \)
• \( \chi_m = 2 \times 10^{-5}. \)

 

Rearranging the formula for \( H \):

\[ H = \frac{M}{\chi_m}. \]

Substitute the given values:

\[ H = \frac{5 \times 10^{-2}}{2 \times 10^{-5}} = 2500 \, \text{A/m}. \]

Step 2: Relate the magnetic field strength \( H \) to the geometry of the toroid.

The magnetic field strength \( H \) inside a toroid is given by:

\[ H = \frac{N I}{2 \pi r}, \]

where:

• \( N \) is the number of turns of wire,
• \( I \) is the current through the wire, and
• \( r \) is the radius of the toroid's core (in meters).

 

We are given:

• \( N = 500, \)
• \( I = 2\pi \, \text{A}, \)
• \( H = 2500 \, \text{A/m}. \)

 

Rearranging the formula for \( r \):

\[ r = \frac{N I}{2 \pi H}. \]

Substitute the values:

\[ r = \frac{500 \cdot 2\pi}{2 \pi \cdot 2500}. \]

Simplify:

\[ r = \frac{500}{2500} = 0.2 \, \text{m} = 20 \, \text{cm}. \]

Final Answer: The radius of the ring is \( \mathbf{20 \, \text{cm}} \), which corresponds to option \( \mathbf{(D)} \).