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Step-by-Step Solution
Step 1: Understand the Problem
We have an iron rod of volume 10−3 m3 (relative permeability 1000) acting as the core of a solenoid with 10 turns/cm. A current of 0.5 A flows through the solenoid. We need to find the magnetic moment of the iron rod.
Step 2: Convert Units and Enumerate Known Values
Volume of the rod, V = 10−3 m3
Relative permeability, $ \mu_r $ = 1000
Current, I = 0.5 A
Turns per cm for the solenoid is 10 turns/cm. Converting to turns per meter:
10 turns/cm = 10 ÷ (10−2 m) = 1000 turns/m
Step 3: Express Volume in Terms of Cross-sectional Area and Length
Since the rod’s volume is V = 10−3 m3, we can denote it as
$ V = A \times l $,
where A is the cross-sectional area and l is the length of the rod. We will use this expression in the formula involving A and l.
Step 4: Write Down the Formula for Magnetic Moment
The magnetic moment M for a rod placed inside a solenoid can be written in one common form as:
$$ M = (n \, l) \times I \times A \, (\mu_r - 1), $$
where
n is the number of turns per unit length of the solenoid,
l is the length of the rod,
I is the current through the solenoid,
A is the cross-sectional area of the rod,
$\mu_r$ is the relative permeability of the rod.
Because A l = V, we can substitute V directly into the expression.
Step 5: Substitute Numerical Values
Using n = 1000 turns/m, I = 0.5 A, V = 10−3 m3, and $ \mu_r = 1000 $:
$$
M = n \, I \, (A \, l) \, (\mu_r - 1)
= 1000 \times 0.5 \times 10^{-3} \times (1000 - 1).
$$
Step 6: Perform the Calculation
Inside the parentheses, $ (1000 - 1) = 999 $.
Thus,
$$
M = 1000 \times 0.5 \times 10^{-3} \times 999
= (1000 \times 10^{-3}) \times (0.5 \times 999 ).
$$
First, $ 1000 \times 10^{-3} = 1 $. Therefore,
$$
M = 1 \times (0.5 \times 999) = 499.5 \, \text{A m}^2.
$$
Since 499.5 is approximately 500, we can write:
$$
M \approx 5 \times 10^2 \, \text{A m}^2.
$$
Step 7: Final Answer
The magnetic moment of the rod is
$$ 5 \times 10^2 \,\text{A m}^2. $$
