Consider using the torque on a current-carrying coil in a magnetic field to detect relatively small magnetic fields (less than the field of the Earth, for example). Construct a problem in which you calculate the maximum torque on a current-carrying loop in a magnetic field. Among the things to be considered are the size of the coil, the number of loops it has, the current you pass through the coil, and the size of the field you wish to detect. Discuss whether the torque produced is large enough to be effectively measured. Your instructor may also wish for you to consider the effects, if any, of the field produced by the coil on the surroundings that could affect detection of the small field.

Magnetic field lines are never crossed. The strength of the field is determined by the density of the field lines.

Calculate the maximum torque on a\({\rm{200}}\)turn

The radius of coil =\({\rm{15}}\;{\rm{cm}}\left( {\frac{{{\rm{1}}\;{\rm{m}}}}{{{\rm{100}}\;{\rm{cm}}}}} \right) = {\rm{0}}{\rm{.15}}\;{\rm{m}}\)

Current=\({\rm{100}}\;{\rm{\mu A}}\left( {\frac{{{{10}^{ - 6}}\;{\rm{A}}}}{{{\rm{1}}\;{\rm{\mu A}}}}} \right) = {\rm{1}}{\rm{.00}} \times {10^{ - {\rm{4}}}}\;{\rm{A}}\)

Magnetic Field = \({\rm{3}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}\;{\rm{T}}\)

(a)

The formula \(\tau = {\rm{NIABsin\theta }}\) can be used to calculate torque. By replacing the variables and remembering that the highest torque occurs when \({\rm{sin\theta = 1}}\), we get

\(\begin{aligned}{}\tau &= {\rm{NIAB}}\\ &= {\rm{200}} \times \left( {{\rm{1}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{A}}} \right) \times {{\rm{(0}}{\rm{.15}}\;{\rm{m)}}^{\rm{2}}} \times {\rm{\pi }} \times \left( {{\rm{3}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}\;{\rm{T}}} \right)\\ &= {\rm{4}}{\rm{.24}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 10}}}}\;{\rm{N}} \cdot {\rm{m}}\end{aligned}\)

This is a very small amount of torque that can't be measured with a traditional dynamometer like those used in classroom demonstrations. To accurately detect minuscule magnetic fields with such small torques, significantly more complex torque-measuring instruments and methodologies would be required.

Therefore, torque is \({\rm{4}}{\rm{.24}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 10}}}}\;{\rm{N}} \cdot {\rm{m}}\).

(b)

The coil in question produces a magnetic field that is

\(\begin{aligned}{}{\rm{B}} &= \frac{{{\rm{N}}{{\rm{\mu }}_{\rm{0}}}{\rm{I}}}}{{{\rm{2R}}}}\\ &= \frac{{{\rm{200}} \times \left( {{\rm{4\pi }} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}\frac{{{\rm{Tm}}}}{{\rm{A}}}} \right) \times \left( {{\rm{1}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{A}}} \right)}}{{{\rm{2}} \times {\rm{0}}{\rm{.15}}\;{\rm{m}}}}\\ &= {\rm{8}}{\rm{.38}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 8}}}}\;{\rm{T}}\end{aligned}\)

This field is one order of magnitude less than the external field. Hence, it might cause slight interference and detection mistakes.

Therefore, the magnetic field is \({\rm{8}}{\rm{.38}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 8}}}}\;{\rm{T}}\).