Question

Are circular coil of radius 40mm is found to produce a field 12.1×10^-6 T at a point 30mm along its axis.(a) what is the electromagnetic moment? (b) what maximum torque could it experience in a uniform magnetic field of 3.0×10^-3T.

Answer 1

a) Electromagnetic moment of the coil is given as:

$$m = IS$$

where $S = \pi R^2$ is the area of the coil, where $R = 40mm = 0.04m$ is its radius, and $I$ is the current in the coil. In turn the current can be found from the following expression:

$$B = \dfrac{\mu_0I}{2R}\left( 1+\dfrac{h^2}{R^2} \right)^{-3/2}$$

where $B = 12.1\times 10^{-6}T$ is the bagnetic field, $h =30mm = 0.03m$ is the distance along the axis, and $\mu_0 =4\pi\times 10^{-7}H/m$ is the magnetic permeability. Expressing the current, obtain:

$$I = \dfrac{2RB}{\mu_0}\left( 1+\dfrac{h^2}{R^2} \right)^{3/2}$$

And the moment:

$$m = \dfrac{2\pi R^3B}{\mu_0}\left( 1+\dfrac{h^2}{R^2} \right)^{3/2}\\ m = \dfrac{2\pi \cdot 0.04^3\cdot 12.1\times 10^{-6}}{4\pi\cdot 10^{-7}}\left( 1+\dfrac{0.03^2}{0.04^2} \right)^{3/2} \approx 0.0076\space A\cdot m^2$$

b) The maximum torque is given as follows:

$$\tau = mB_0$$

where $B_0 = 3\times 10^{-3}T$ is the external field. Thus, obtain:

$$\tau \approx 2.3\times 10^{-5}\space N\cdot m$$

Answer. a) $0.0076\space A\cdot m^2$ b) $2.3\times 10^{-5}\space N\cdot m$.