The right edge of the circuit extends into a 50 mT uniform magnetic field. What

Question

The right edge of the circuit extends into a 50 mT uniform magnetic field. What are the magnitude and direction of the net force on the circuit? (0.025 N) The rectangular loop shown has a mass of 0.15g per centimeter of length and is pivoted about side ab on a frictionless axis. The current in the wire is 8.2 A in the direction shown. Find the magnitude and direction of the magnetic field parallel to the y-axis that will cause the loop to swing up until it makes an angle of 30degree with the yz-plane. (0.0242 T in the + y-dir)

Explanation / Answer

1. I= 15v/3.0 Ohms= 5v

the magnitude of force is F=ILB = 5 x.10 m x .05 T= .025 N

Direction of force is upwards.

2. Equating the torque of the magnetic field to the TORQUE of gravity.

Tb=IBAsin(60)

Tg=r x Fg --> (r=9/2cm = 4.5cm) because Fg should be centered on the plane of the loop
m=.00015kg*30 = 0.0045kg

Tg=(0.045m)*(0.0045kg)(9.8)*sin(150) angle btw Fg and plane of loop
Tg=0.00099225 Nm

Equate Tg and Tb
0.00099225=IBAsin(60) --> 0.00099225=(8.2A)B(0.08m*0.06m)sin(60) --> B=0.0242 T in +y direction

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