A particular Alnico (aluminum, cobalt, nickel, and iron) bar magnet (magnet A) has a mass of 10 g. It produces a magnetic field of magnitude 3e-05 T at a location 0.17 m from the center of the magnet, on the axis of the magnet. (a) Approximately what is the magnitude of the magnetic field of magnet A a distance of 0.51 m from the center of the magnet, along the same axis

The magnet applies a magnetic field 0.7 * 10^-5 T at distance 0.51 m

Explanation:

Solution

The magnetic field is produced due to the orientation of the electrons in their orbits in the material of the magnet where the bar magnet is a magnetic dipole where the magnetic field produced due to the magnet is given by

B_a=μ_o*2μ/4*π*r^3 (1)

Where μ is the magnetic dipole, the term μ_o/4*π is constant and equals 1 x 10 ^-7 T. m^2/C. m/s and r is the distance from the magnet to the location where the magnetic field is applied. μ is the magnetic dipole.

As shown by equation (1), the magnetic field is inversely proportional to

r_3 (B ∝ 1/r^3). therefore, for two instants r1 and r2, we could get the next r3 relationship in the form

B_1/B_2=r_2^2/r_1^3

B_2 = (r_1^3/r_2^3) B_1 (2)

Where B_1 = 3 x 10^ - 5 T, r_1 = 0.17 m and r_2 = 0.51 m. Now we can plug our values for B_1, r_1 and r_2 into equation (2) to get B_2

B_2 = (r_1^3/r_2^3) B_1

=0.7 * 10^-5 T

The magnet applies a magnetic field 0.7 * 10^-5 T at distance 0.51 m