A circular coil of wire 8.40 in diameter has 18.0 turns and carries a current of

Question

A circular coil of wire 8.40 in diameter has 18.0 turns and carries a current of 2.70 . The coil is in a region where the magnetic field is 0.640 .

1)What orientation of the coil gives the maximum torque on the coil ?
Please, enter the value of the angle between the field and the normal to the plane of the loop.

2)What is this maximum torque in part (1) ?

3)For what orientation of the coil is the magnitude of the torque 71.0 of the maximum found in part (2)?
Please, enter the value of the angle between the field and the normal to the plane of the loop.

Explanation / Answer

Here the torque is given by the vector cross product
T = M x B

M is a vector with direction along the area vector of the coil

magnitude of M = niA = (18) (2.70) ( x 0.107m2) [Radius= 8.4/2 inches= 10.69cm = 0.107m]

= 0.036

B = 0.640 T

1) cross product = M B sin

sin is max for =90 degrees.

So the torgue is max when the magnetic field is along the plane of coil.

2) max torque = M B sin90 =  0.036 x  0.640

= 0.023 Nm

3) Required torque = 0.71 x  0.023 = 0.0164

So MBsin = 0.0164 ie sin = 0.711

So = 45.3 degrees

Hope you got the concept :)

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