The Biot-Savart law gives the contribution of a current element to the
Answer: A
The contribution falls as the inverse square of the distance and is directed at right angles to both the element and the line joining it to the point. It plays for magnetostatics the part Coulomb's law plays for electrostatics.
Please log in or register to report this question.
In the Biot-Savart law the contribution of a current element is zero at points
Answer: B
The cross product of the element with the unit vector to the point vanishes when the two are parallel, so a straight wire produces no field at points on its own line extended. The contribution is greatest at right angles to the element.
Please log in or register to report this question.
The direction of the magnetic field round a straight current-carrying conductor is
Answer: C
Gripping the conductor with the right hand so the thumb points along the current makes the fingers curl the way the field goes. The field therefore has no radial or axial component at all, which is what makes a coaxial Amperian loop so effective.
Please log in or register to report this question.
Magnetic field intensity \( \mathbf{H} \) has units of
Answer: D
Since Ampere's law equates the line integral of H to a current, H must be a current divided by a length. Multiplying by the permeability gives the flux density in teslas, which is a very much larger number in free space.
Please log in or register to report this question.
The relation between flux density and field intensity in a linear medium is
Answer: A
The permeability of free space is \( 4\pi \times 10^{-7} \) henries per metre, so a field intensity of one ampere per metre corresponds to a flux density of about 1.26 microteslas. In iron the relative permeability may be several thousand.
Please log in or register to report this question.
The permeability of free space has the value
Answer: B
It was formerly exact by the definition of the ampere, and it combines with the permittivity to give the speed of light as the reciprocal of the square root of their product. That relation is one of the great results of Maxwell's theory.
Please log in or register to report this question.
Ampere's circuital law states that the line integral of \( \mathbf{H} \) round a closed path equals
Answer: C
Written \( \oint \mathbf{H} \cdot d\mathbf{l} = I_{enc} \), it is the magnetic counterpart of Gauss's law and is the quickest route to the field of any arrangement with enough symmetry. In terms of B the permeability would appear as a factor.
Please log in or register to report this question.
For Ampere's law to yield the field directly, the Amperian path must be chosen so that
Answer: D
The dot product then becomes a constant magnitude times a length, and the field can be solved for. Where the field is perpendicular to the path the contribution vanishes, which disposes of the unwanted portions of a rectangular path.
Please log in or register to report this question.
The field at a distance \( \rho \) from a long straight wire carrying current \( I \) is
Answer: A
A circular Amperian path of circumference \( 2\pi\rho \) encloses the whole current, so the field intensity is the current over that circumference. Multiplying by the permeability gives the flux density, which is the fourth option.
Please log in or register to report this question.
Inside a solid conductor of radius \( a \) carrying a uniformly distributed current, the field at radius \( \rho \) varies as
Answer: B
The enclosed current grows as the square of the radius while the path length grows as the first power, so their ratio rises linearly. The field is therefore greatest at the surface and falls as the inverse first power beyond it.
Please log in or register to report this question.
