Question #196040

(a) What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? _____ N-m

(b) What is the torque when θ is 10.9°? _____ N-m


Round your answers to the nearest 1 decimal place

1
Expert's answer
2021-05-28T07:18:50-0400

Explanations & Calculations


  • The Torque generated by a loop of a conductor placed in a uniform magnetic field is given by

τ=BiNAsinθ\qquad\qquad \begin{aligned} \small \tau&=\small BiNA\sin\theta \end{aligned}

  • The angle is measured with respect to the plane of the magnetic field such that at 90 degrees the torque achieves its maximum as sin90=1\small \sin 90 = 1.
  • That is when the loop positions itself perpendicular to the field.
  • Similarly, no torque is experienced when the loop is parallel to the field when θ=0\small \theta = 0 as sin0=0\small \sin 0 =0.
  • Then,

τmax=BiNA=1.60T×50A×150×(0.18m)2=388.8Nmτ@θ=10.90=1.60×50×150×(1.8)2×sin10.9=73.5Nm\qquad\qquad \begin{aligned} \small \tau_{max}&=\small BiNA\\ &= \small 1.60T\times50A\times150\times(0.18m)^2\\ &=\small \bold{388.8\,Nm}\\\\ \tau_{@\,\theta=10.9^0}&=\small 1.60\times50\times150\times(1.8)^2\times\sin10.9\\ &=\small \bold{73.5\,Nm} \end{aligned}


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