Question

Two identical small current carrying loops of wire are oriented along a common axis with the plane of their loops parallel to each other. See Figure EC. The distance between the loops, x, is much larger than the radius of each loop, R. The magnetic dipole moment of each loop is directed parallel to the axis, and the magnetic field that one loop creates at the center of the other loop is also parallel to the axis. So mu^rightarrow and B^rightarrow are parallel to each other. Therefore the magnetic potential energy of the system can be written: U_B = -mu^rightarrow middot B^rightarrow = -muB. Note that B is a function of x, the distance between the 2 loops. Lastly, the magnitude of the magnetic force that each loop exerts on the other one can be found in the following way: |F| = |dU/dx| = |mu|dB\dx What is the value (magnitude) of the magnetic dipole moment of a single loop of wire with radius R that carries a current I? Find the magnitude of the magnetic field created by the current in one loop at the center point of the other loop. (Find B in terms of I, x, R and known constants). Use the formula given above to find the magnitude of the magnetic force of one loop on the other loop. Show that if x NestedGreaterGreater R, then this force is proportional to 1/x^4. Evaluate this force for x = 15.0 cm, R = 0.4 cm, and I = 8.0 A. (Each loop has the same current.)

Answer 1

a) By definition: $\mu =IA=I\cdot \pi R^{2}$ (it is a vector, but here is the formula for its modulus)

b) $B=\frac{\mu _{0}R^{2}I}{2\left ( x^{2} +R^{2}\right )^{3/2}}$ (there is a deduction; ife necessary, I'll give it here)

$\mu _{0}$ =permitivity of vacuum

c) Use the formula for F given in the text, and get:

$\left | F \right |=\frac{3\pi \mu _{0}R^{4}I}{2}\cdot \frac{x\sqrt{x^{2}+R^{2}}}{\left ( x^{2} +R^{2}\right )^{3}}\: \left ( 1 \right )$

d) Extract x2 from the radical at numerator, and x2 - forced factor from denominator. We'll get:

$\left | F \right |=\frac{3\pi \mu _{0}R^{4}I}{2}\cdot \frac{x^{2}\sqrt{1+\frac{R^{2}}{x^{2}} }}{\left [x^{2} \left ( 1+\frac{R^{2}}{x^{2}} \right )\right ]^{3}}\sim \frac{1}{x^{4}}$

[the radical from the numerator=1; the small bracket from the denominator=1]

d) Replace the numbers in (1) and get: F~2.3914*10^-13 N.