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Answer: The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center. So it would be D.
Answer: D
Explanation:
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The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
