Question

A wire is formed into a circle having a diameter of 10.3 cm and is placed in a uniform magnetic field of 2.98 mT. The wire carries a current of 5.00 A. Find the maximum torque on the wire.

Answer 1

Answer:

T(max) = 1.17 × 10⁻⁴Nm

= 117μNm

Explanation:

T = BIA sinθ

A = area enclosed

θ = angle between normal plane

for max. torque θ = 90, (sin90° =1)

T = BIA sin90°

T= BI (πd/4)

T = $T_m_a_x = \frac{1}{4} (2.98 * 10^-^3)(5)(\pi )\\T_m_a_x = 1.17 * 10^-^4Nm\\T_m_a_x = 117UNm$