A vacuum chamber in the basement of Kettering University is filled with a unifor

Question

A vacuum chamber in the basement of Kettering University is filled with a uniform magnetic field of magnitude  6.4×102T pointing in the negative y -direction. Choose your coordinate system so that the origin coincides with the floor of the chamber and the positive z -direction points vertically upward. Laying on the floor of the chamber are two cylindrical iron rods (one long and one short) laying parallel to the x -axis. The long iron rod has a length of  1.91m and a radius of 2.7×103m , while the short iron rod has a length of  4×102m and a radius of 2.7×103m . The centers of the two rods are separated by a distance of 6×102m . The resistivity of iron is 9.7×108m.

The long rod remains connected to the first battery, and at t=  22s another student connects the ends of the short rod to the terminals of a 230V  battery so that a conventional current flows through it in the positive x -direction.

What is the magnitude of the net magnetic force exerted on the short rod immediately after connecting it to the second battery?

Explanation / Answer

Resistance on each rod R = rho L/A

rho is ressitivity and A is area = pi r^2

R1 = 9.7 *10^-8 * 1.91/(3.14* 0.0027^2)

R1 = 0.00809 ohms

Current I = I1 = I2 = V/R

I1 = 230/0.00809

I1 = I2 = 2.84 *10^4 Amps

V is Potential Difference

Force F = ILB sin theta

F1 = 2.84*10^4 * 1.91* 6.4*10^-2* sin 90

F1 = 3471.61 N

L is length of the rod

force on second is contibute by both B and B1

B1 = uoI1/(2pir)

B1 = (4pi *10^-7 * 2.84 *10^4)/(2* 3.14 *0.06)

B1 = 0.0947 T

as this is opposite to B ,

B2 = B - B1

B2 = 0.064 - 0.0947

B2 = 0.0307T

finally F2 = I2 L B2

F2 = (2.84 *10^4 * 1.91* 0.0307)

F2 = 1.66*10^3 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.