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QPages 286-287: Q3: HC Verma: Concepts of Physics - Volume 2 Magnetic Properties of Matter Pages 286-287: Q3 Solution: The magnetic field inside a long solenoid of 50 turns is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly equate the total magnetic field $B$ with magnetization $I$ or fail to subtract the air-core contribution $mu_0 H$ from the total field before dividing by $mu_0$. Remember that $B = mu_0(H + I)$, so you must isolate $I$ by recognizing that $mu_0 H$ is the field produced by the solenoid alone.

The magnetic field inside a long solenoid of 50 turns cm−1\text{cm}^{-1} is increased from 2.5×10−3 T2.5 \times 10^{-3} \text{ T} to 2.5 T when an iron core of cross-sectional area 4 cm24 \text{ cm}^2 is inserted into it. Find a) the current in the solenoid b) the magnetisation I of the core and c) the pole strength developed in the core.

Step-by-Step Verified Solution

**Answer:** (a) i≈0.4 Ai \approx 0.4 \text{ A} (b) I≈2×106 A m−1I \approx 2 \times 10^6 \text{ A m}^{-1} (c) m=800 A mm = 800 \text{ A m}. **Step-BY-Step Explanation:** Given: n=50 turns/cm=5000 turns/mn = 50 \text{ turns/cm} = 5000 \text{ turns/m}, B1=2.5×10−3 TB_1 = 2.5 \times 10^{-3} \text{ T} (without core), B2=2.5 TB_2 = 2.5 \text{ T} (with core), A=4×10−4 m2A = 4 \times 10^{-4} \text{ m}^2. (a) Without core: B1=μ0niB_1 = \mu_0 n i. i=B1μ0n=2.5×10−34π×10−7×5000≈0.4 Ai = \frac{B_1}{\mu_0 n} = \frac{2.5 \times 10^{-3}}{4\pi \times 10^{-7} \times 5000} \approx 0.4 \text{ A}. (b) With core: B2=μ0(H+I)=μ0H+μ0IB_2 = \mu_0 (H + I) = \mu_0 H + \mu_0 I. Since H=ni=5000×0.4=2000 A/mH = ni = 5000 \times 0.4 = 2000 \text{ A/m}, μ0H=2.5×10−3 T\mu_0 H = 2.5 \times 10^{-3} \text{ T}. So μ0I=2.5−0.0025≈2.5 T\mu_0 I = 2.5 - 0.0025 \approx 2.5 \text{ T}. I=2.54π×10−7≈2×106 A/mI = \frac{2.5}{4\pi \times 10^{-7}} \approx 2 \times 10^6 \text{ A/m}. (c) Pole strength m=I×A=2×106×4×10−4=800 A mm = I \times A = 2 \times 10^6 \times 4 \times 10^{-4} = 800 \text{ A m}.

All Questions in this Chapter

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