Q.23 A current carrying toroid winding is internally filled with lithium having susceptibility chi = 2.1×10⁻⁵ . What is the percentage increase in the magnetic field in the presence of lithium over that without it?

Answer: Percentage increase = 2.1×10⁻³ % Without lithium: B₀ = μ₀ H . With lithium filling the toroid: B = μ H = μ₀(1 + chi)H . Fractional increase: B - B₀/B₀ = μ₀(1+chi)H - μ₀H/μ₀H = chi . Percentage increase = chi×100 = 2.1×10⁻⁵×100 = 2.1×10⁻³ %. (Lithium is paramagnetic, so the field increases, but only very slightly.)

Chapter: Magnetic Materials· Physics

A current carrying toroid winding is internally filled with lithium having susceptibility \(\chi = 2.1\times10^{-5}\). What is the percentage increase in the magnetic field in the presence of lithium over that without it?

Q.23 A current carrying toroid winding is internally filled with lithium having susceptibility \(\chi = 2.1\times10^{-5}\). What is the percentage increase in the magnetic field in the presence of lithium over that without it?

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: Percentage increase \(= 2.1\times10^{-3}\) %

Without lithium: \(B_0 = \mu_0 H\). With lithium filling the toroid: \(B = \mu H = \mu_0(1 + \chi)H\).

Fractional increase: \(\dfrac{B - B_0}{B_0} = \dfrac{\mu_0(1+\chi)H - \mu_0H}{\mu_0H} = \chi\).

Percentage increase \(= \chi\times100 = 2.1\times10^{-5}\times100 = 2.1\times10^{-3}\) %.

(Lithium is paramagnetic, so the field increases, but only very slightly.)

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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A current carrying toroid winding is internally filled with lithium having susceptibility \(\chi = 2.1\times10^{-5}\). What is the percentage increase in the magnetic field in the presence of lithium over that without it?

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