PROBLEM 6.1KNOWN:A current loop having N = 20 A = 1 in2= 0.000645 m2 I = 0.02 A B = 0.4 Wb/m2FIND:Torque on the current loop, Tµ. ASSUMPTIONS:The magnetic field is oriented at an angle of 90oto the current flow direction. SOLUTION:From (6.3) TNIABµα=sinThe maximum torque occurs when sin α= 1, which yields ()()()max22200.02 A0.000645 m0.4 Wb/mTNIABµ==max41.0310N-mTµ−=×
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