A square metal loop with resistance R = 3 and sides a = 10cm is rotating in a un

Question

A square metal loop with resistance R = 3 and sides a = 10cm is rotating in a uniform magnetic field B = 5mT pointing out of the page. The angle between B and A (the area vector of the loop) is changing with time as the loop rotates. The angle is given by the function q =t4 where a is a constant with appropriate units.

WE KNOW THE FOLLOWING:

 

**Answer these questions for the instant of time shown in the diagram.**

i.) (2pts) Find the magnitude of the force FInd induced on any of the sides of the loop. Give an expression in terms of a, B, R, , , and t .

 

|FInd | = _________________

 

j.) (2pts) Find the magnitude of the magnetic dipole moment m induced in the loop. Give an expression in
terms of a, B, R, , , and t .
| m|= _________________

 

k.) (1pt) What is the direction of the magnetic dipole moment at the instant in the diagram? (Circle one.)

 

1.Clockwise around the loop
2.Counter-Clockwise around the loop
3.Normal to the loop inward
4.Normal to the loop outward

 

l.) (2pts) Find the magnitude of the torque induced on the loop. Give an expression in terms of a, B, R, , , and t . (Include units.)
||= _________________

 

I HAVE THE ANSWERS BUT I DO NOT KNOW HOW TO SOLVE THEM.

Answers:

i.)|FInd|= 4t3B2a3 sin/(R)

j.)| m|= 4t3Ba3 sin/(R)

k.) 4.Normal to the loop outward

l.)||= (4t3Ba3 sin)/(R)

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Please show your work for all of them. Make sure is alpha AND a is "a"

Explanation / Answer

the magnitude of the force is                 Find = IlB sin      (since , l = a) the angle between l = a and magnetic field B is = 900 since, sin90 = 1               Find = IlB       since, sin90 = 1               Find = IlB                              = (4t3Ba2sin / R)(a)(B)                        = 4t3B2a3sin / R _______________________________________ _______________________________________ the magnitude of the magnetic dipole moment is             m = NAI for single loop N = 1 The area of the square loop = 2a from axis of symmetry (due to rotation) A = 2a/2 = a               m = NAI                   = (1)(a)(4t3Ba2sin / R)                   = 4t3Ba3sin / R __________________________________________ __________________________________________ the magnitude of the torque induced on the loop is                = IAB                  = (4t3Ba2sin / R)(a)(B)                  = 4t3B2a3sin / R

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