Flux passing from air into iron is bent
Answer: C
The tangential intensity is preserved so the tangential flux density is multiplied by the permeability, while the normal flux density is unchanged. In iron of relative permeability a thousand, flux entering at any appreciable angle emerges running almost parallel to the surface.
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Flux emerging from iron into air is very nearly
Answer: D
The relation between the two angles has the tangent in air smaller by the ratio of the permeabilities, so a wide range of angles within the iron corresponds to a narrow cone about the normal outside. This is why the iron surface behaves almost like a magnetic equipotential.
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A charge of 2 \( \mu \)C moves at \( 10^{5} \) m/s perpendicular to a field of 0.5 T. The force on it is
Answer: A
Correct answer: 0.1 N
\[ F = qvB\sin\theta = 2\times10^{-6} \times 10^{5} \times 0.5 = 0.1\ \mathrm{N}, \] with the sine equal to one since the motion is perpendicular to the field. The force acts at right angles to both the velocity and the field, so it changes the direction of motion without doing any work - which is why a magnetic field alone cannot alter a particle’s speed.
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An electron moving at \( 10^{7} \) m/s enters a field of 0.01 T at right angles. The radius of its path is about
Answer: B
Substituting in \( r = mv/(QB) \) gives \( 9.11 \times 10^{-31} \times 10^{7}/(1.6 \times 10^{-19} \times 0.01) = 5.7 \times 10^{-3} \) m. A field of only a hundredth of a tesla therefore bends a fast electron into a path of a few millimetres.
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A conductor 0.5 m long carries 20 A at right angles to a field of 0.8 T. The force on it is
Answer: C
Correct answer: 8 N
\[ F = BIL\sin\theta = 0.8 \times 20 \times 0.5 = 8\ \mathrm{N}. \] It is the same force as on the previous item’s moving charge, summed over all the charges in the conductor, which is why the two expressions have the same form. The force is perpendicular to both the conductor and the field, and it is what turns every electric motor.
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That same conductor is turned so that it lies at 30 degrees to the field. The force becomes
Answer: D
The force is proportional to the sine of the angle between the conductor and the field, and the sine of thirty degrees is one half, so the force is halved to four newtons. At zero degrees, along the field, the force would vanish altogether.
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Two long parallel conductors 0.1 m apart carry 100 A each in opposite directions. The force per metre is
Answer: A
The expression gives \( 2 \times 10^{-7} \times 10^{4}/0.1 = 2 \times 10^{-2} \) N/m, and opposing currents repel. On a short circuit of ten kiloamperes the same conductors would push apart with two hundred newtons per metre, which is why busbar bracing matters.
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A loop of area \( 0.02 \) m² carrying 5 A lies in a field of 0.4 T with its plane containing the field. The torque on it is
Answer: B
The moment is \( 5 \times 0.02 = 0.1 \) A·m², and multiplying by the flux density with the sine of ninety degrees gives 0.04 newton metres. When the plane contains the field the moment is perpendicular to it and the torque is greatest.
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A material of relative permeability 500 has a magnetic susceptibility of
Answer: C
Since \( \mu_r = 1 + \chi_m \), the susceptibility is one less than the relative permeability. For a ferromagnetic material the difference of one is negligible, but for a diamagnetic or paramagnetic material the susceptibility is the only quantity worth quoting.
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A specimen in a field intensity of 1000 A/m has a magnetisation of 500 kA/m. Its flux density is about
Answer: D
Correct answer: 0.63 T
\[ B = \mu_0(H+M) = 4\pi\times10^{-7} \times (1000 + 500{,}000) = 1.257\times10^{-6} \times 5.01\times10^{5} = 0.63\ \mathrm{T}. \] The magnetisation exceeds the applied field intensity five hundredfold, so it contributes essentially all the flux density - which is the whole point of a ferromagnetic core, and why omitting \( M \) would give only 1.26 mT.
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