Question

1.  A rectangular loop of wire (0.2 m wide and 0.3 m tall) carrying a current of 0.6 A is in the xy plane, with its right edge along the +y axis. A magnetic field (uniform, B = 1.2 T) exists along a direction that is 74° from the +z axis in the xz plane. What magnitude of torque will act on the loop of wire at that moment ? a) 0.42 Nm b)0.15 Nm c) 0.04 Nm d) 0.09 Nm

2. If the B-field in the previous problem was oriented at 50° from +x in the xz plane determine the magnitude and direction of the torque on a 150 turn loop of the same size as the coil in question#1. a) 4.2 Nm, along +y b) 2.8 Nm, along +y c) 1.1 Nm, along +z d) 0.5 Nm, along +z

3. Will the plane of the coil of wire in question #2 initially swing toward the +z or the -z axis? a) +z b) -z c) Neither +z nor -z

4. For the torque on a current carrying loop to be maximized, the angle between the normal to the plane of the loop and the B-field must be: a) zero. b)45°. c) 90° d) 180°

Answer 1

Torqye on a current loop in a magnetic field is given by

  

Where, I is current in the loop

A is area of the loop

B is magnetic field

N is number of turns in the loop

(1) Current in the loop = 0.6 A

Area of the loop = 0.2 x 0.3 = 0.06 m²

Angle between the area vector and the magnetic field = 74⁰

  Number of turns in the loop = 1

Magnetic field strength = 1.2 T

So, torque on the loop is

  

Answer : (c)

(2) Number of turns in the loop = 150

Angle between area vector (normal to te plane of loop) and magnetic field = 90 -50 = 40^o

So, torque on the loop is

  

Answer : (a)

(3) By using right hand screw rule, we can determine direction of the torque. In the given problem direction of the torque is towards the y - direction.

As, magnetic field is in xz - plane, direction of area vector is along the z - direction. So, cross product (torque ) must be along the y - direction.

Answer : (c)

(4)

For maximum torque the value of should be maximum. Therefore should be 90^o .

Answer : (c)

For any doubt please comment.