Solution

(a) The magnitude of the toroidal field is given by $$B_0 = μ_0ni_p$$, where n is the number of turns per unit length of the toroid, and $$i_p$$ is the current required to produce the field (in the absence of the ferromagnetic material). We use the average radius ($$r_{avg} = 5.5$$ cm) to
calculate n:

$$n=\dfrac{N}{2\pi r_{avg}}=\dfrac{400\,turns}{2\pi(5.5 \times 10^{-2})}=1.16 \times 10^{3\,}\,turns/m$$

Thus,

$$i_p=\dfrac{B_0}{\mu_0 n}=\dfrac{0.20 \times 10^{-3}\,T}{(4\pi \times 10^{-7}\,T.m/A)(1.16 \times 10^{3}/m)}=0.14\,A$$

(b) If Φ is the magnetic flux through the secondary coil, then the magnitude of the emf induced in that coil is $$ε = N(dΦ/dt)$$ and the current in the secondary is $$i_s = ε/R$$, where R is the resistance of the coil. Thus,

$$i_s=\Bigg(\dfrac{N}{R}\Bigg)\dfrac{d\phi}{dt}$$

The charge that passes through the secondary when the primary current is turned on is

$$q=\int i_s\,dt=\Bigg(\dfrac{N}{R}\Bigg)\int \dfrac{d\phi}{dt}=\Bigg(\dfrac{N}{R}\Bigg)\int^\phi_0 d\phi=\dfrac{N\phi}{R}$$

The magnetic field through the secondary coil has a magnitude $$B = B_0 + B_M = 801B_0$$,where $$B_M$$ is the field of the magnetic dipoles in the magnetic material. The total field is perpendicular to the plane of the secondary coil, so the magnetic flux is $$Φ = AB$$, where A is the area of the Rowland ring (the field is inside the ring, not in the region between the
ring and coil). If r is the radius of the ring’s cross-section, then $$A = πr^2$$. Thus,

$$Φ = 801\pi r^2B_0$$

The radius r is $$(6.0 \,cm – 5.0\, cm)/2 = 0.50$$ cm and

$$Φ = 801\pi(0.50\times10^{-2}\,m)^2(0.20 \times 10^{-3}\,T)=1.26 \times 10^{-5}\,Wb$$

Consequently, $$q=\dfrac{50(1.26 10 Wb)}{8.0\,\Omega}=7.9 \times 10^{-5}\,C$$.