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Chapter 4: Q74P (page 918)

The lower end of the thin uniform rod in Figure is attached to the floor by a frictionless hinge at point P. The rod has mass 0.0840kg and length 18.0cm and is in a uniform magnetic field B = 0.120T that is directed into the page. The rod is held at an angleu=53掳 above the horizontal by a horizontal string that connects the top of the rod to the wall. The rod carries a current I = 12.0 A in the direction toward P. Calculate the tension in the string. Use your result from Problem 27.70 to calculate the torque due to the magnetic-field force.

Short Answer

Expert verified

The tension in the string is 0.472N

Step by step solution

01

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

02

Determine the torqueThe tension in the string can be calculated by the relation

T=mgcos鈦�(胃)+IBL2sin鈦�(胃)T=(0.0840kg)9.80m/s2cos鈦�53鈭�+(12.0A)(0.120T)(0.180m)2sin鈦�53鈭�T=0.472N

Hence, the tension in the string is 0.472N.

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