Aflat, round iron ring 5.80cm in diameter has a current running through it that produces a magnetic field of 77.6? t at its center. this ring is placed in a uniform external magnetic field of 0.380t .

a) what is the maximum torque the external field can exert on the ring? (in n*m)

? max

b) how should the ring be oriented relative to the field for the torque to have its maximum value?

a. the plane of the ring should be parallel to the field

b. the plane of the ring should be normal to the field

A) τmax = 3.59×10^-3 Nm

B) a. the plane of the ring should be parallel to the field

Explanation:

A) the torque exerted by an external magnetic field is given by:

τ = μ×Bext

where μ is perpendicular to the plane of the current loop and has magnitude IA. I is the current in the loop and A is the loop's area.

A circular current loop with radius R has a magnetic field at its center given by B = μ0I/(2R)

I = 2RB/μ0

 = [2(0.0290)(77.6×10-6)]/(4π×10^-7)

 = 3.58A

τ = μ×Bextsin(Ф)

  = IA×Bextsin(Ф)

where Ф is the angle between μ and Bext. the maximum magnitude will occur when sinФ = 1 , Ф = 90°. then:

τmax = Iπ(R^2)Bext

         = (3.58)π((0.029)^2)(0.38)

         = 3.59×10^-3 Nm

b) in order for the ring to experience the maximum torque, the magnetic moment of the loop, μ , and the external magnetic field , Bext, must be perpendicular to each other .

a. the plane of the ring should be parallel to the field

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//