Question

Calculate the magnetic field strength needed on a 449-turn square loop 15 cm on a side to create a maximum torque of 322 N⋅m if the loop is carrying 25 A.

answer to 2 decimal places

Answer 1

Since the magnitude of the torque on a coil of N loops each has the area A, carries a current I and its area vector makes an angle θ with a magnetic field of density B is given by the equation:

$|\vec{τ}|=NIABsinθ$

Therefore, the maximum value of the torque is when θ = 90°, with the magnitude of the torque given by:

$|\vec{τ}_{max}|=BIAN$

From this relation we get that:

$B = \frac{|\vec{τ}_{max}|}{IAN} \\ = \frac{322}{25 \times 0.15^2 \times 449} \\ = 1.27 \;T$