Question

In an attempt to minimize the ice formation on basketball hoops used during Spring Break you decide to run a current carrying wire part way (330 degrees) around the hoops. Assume the current is 300A and the radius of the rim is 30 cm.

A) Derive an equation for the magnetic field at the center of the hoop( show logic not just an integral). Put in numbers to find B(hoop) and direction.

B) If the Earth’s magnetic field is .5 X 10^-4 T and points horizontally to the north on our basketball court, what is the direction and size of the total B vector at the center of the hoop?

C) Determine the force felt by a basketball carrying a charge of 10 coulombs travelling horizontally to the North at 30 m/s just above the center of the rim. Assume field size is the same as at center. Ignore Gravity.

Answer 1

A)Magnetic field dB for a representative piece of wire = dB = I dr x r /4r²

r =R and drxr = k

on integrating both sides

B = I/ 4R^{2 * dl K}

^{=> B =} I s / 4R² where s = arc length

s = R

B =  I / 4R

= 330 * /180 = 5.76 rad

B = 5.76 * 10^-4 T

depends on whether the current is moving clockwise or anti clockwise around the rim

if the current is moving clockwise then the magnetic field moves inward otherwise it moves outwards from the plane of the ring

B)

if the direction of bothe the field is same then

B =  5.76 * 10^-4 + 0.5* 10^-4 = 6.26 * 10^-4 T towards North

if the direction of bothe the field is opposite then

B =  5.76 * 10^-4 - 0.5* 10^-4 = 5.26 * 10^-4 T towards North

C)

Since the magnetic field and the velocity is in perpendicular direction

and magnitude of force = q (vxB) = qvBsin = 0 since =90

F =10 * 30 * 5.26 * 10^-4 = 0.1578 N