Vertically Oriented Dipole magnetism

[dp182]

Now assume you are taking a profile in the x-direction along y=0. Show that the distance between the zero crossings of Bz for this vertical dipole is equal to 2d√2
2, and that the distance between maximum and
minimum values of Bx is equal to d

Homework Equations

B_x=[μ_o/(4∏((x²_p+y²_p+d²)^3/2)](-3dx_p/(x²_p+y²_p+d²))
r_p=[x_p,y_p,0] position of observer
r_q=[0,0,d] vertically downward dipole
x=x_p-x_q

The Attempt at a Solution

I'm not entirely sure how to go about solving this problem any tips on what to start with would be helpful

 

[BigJDubs]

. Assuming the dipole is along the z-direction, the equation for Bz isBz=[μo/(4∏((x2p+y2p+d2)3/2)](3dzp/(x2p+y2p+d2)) At y=0, x=xp so Bz becomes Bz=[μo/(4∏(xp2+d2)3/2)](3dzp/(xp2+d2)) The distance between two zero crossings is equal to 2d√2, therefore, 2d√2 = 2*abs(Bz) and d = abs(Bz). For Bx, the equation is Bx=[μo/(4∏((x2p+y2p+d2)3/2)](-3dxp/(x2p+y2p+d2)) At y=0, x=xp so Bx becomes Bx=[μo/(4∏(xp2+d2)3/2)](-3dxp/(xp2+d2)) Therefore, the distance between maximum and minimum values of Bx is equal to d.