Chapter 6, Problem 6.1P

To determine

The maximum torque on a current loop having 20 turns, a cross-sectional area of 1 in.2 and experiencing a current of 20mA where the magnetic field strength is 0.4 Wb/m².

Answer to Problem 6.1P

The maximum torque on a current loop $T_{\mu }=4.064Nm$.

Explanation of Solution

Given:

Current loop turns = N = $20$

Cross-sectional area = A = $1$ in.2 = $0.0254$ m²(conversion of inch to meter)

Current = I = $20$ mA

Magnetic field strength = B = $0.4$ Wb/m²

Maximum torque = T_µ=O

Calculation:

  $\begin{array}{l}{T}_{\mu }=NIBA\sin \alpha \\ T_{\mu }=20\times 20\times 0.4\times 0.0254\sin {90}^{\circ }\\ T_{\mu }=4.064Nm\end{array}$$\left(\sin {90}^{\circ }=1\right)$

Conclusion:

Thus, the maximum torque on a current loop $T_{\mu }=4.064Nm$.