Question

The mean radius of a Rowland ring is 12 cm and it has 3000 turns of wire wound on its ferromagnetic core of relative permeability 500. If the magnetic field inside the core is 5 T, then the magnetizing current is

• A. 2 A
• B. 3 A
• C. 4 A
• D. 5 A \bigskip

Hint:

A Rowland ring is treated exactly like a toroid. Remember the magnetic field formula \(B=\frac{\mu_0\mu_rNI}{2\pi r}\).

Answer 1

The Correct Option is D

Concept: A Rowland ring behaves like a toroid. For a toroid, \[ B= \frac{\mu_0\mu_rNI}{2\pi r} \] where \[ N=\text{number of turns} \] \[ r=\text{mean radius} \] \[ \mu_r=\text{relative permeability} \] \[ I=\text{magnetizing current} \]

Step 1: Write the toroid magnetic field formula.
\[ B= \frac{\mu_0\mu_rNI}{2\pi r} \] Rearranging, \[ I= \frac{B(2\pi r)} {\mu_0\mu_rN} \]

Step 2: Substitute the given values.
\[ B=5T \] \[ r=12cm=0.12m \] \[ N=3000 \] \[ \mu_r=500 \] \[ \mu_0=4\pi\times10^{-7} \] Therefore, \[ I= \frac{5(2\pi)(0.12)} {(4\pi\times10^{-7})(500)(3000)} \]

Step 3: Simplify the numerator.
\[ 5\times2\pi\times0.12 = 1.2\pi \]

Step 4: Simplify the denominator.
\[ 4\pi\times10^{-7}\times500\times3000 = 0.6\pi \]

Step 5: Calculate the current.
\[ I= \frac{1.2\pi}{0.24\pi} \] \[ I=5A \] Hence, \[ \boxed{I=5A} \] Therefore the correct option is \[ \boxed{\text{(D)}} \]