Correct Answer - Option 3 : 76.4 N-m

**Concept:**

Speed of the motor is given as

\({ω} = \frac{{{N} \times 2\pi }}{{60}}\)

Torque developed by the motor is

T = P / ω

where,

Power developed P = E I_{a}

\(E = \frac{ϕ ZNp}{60A}\)

∴ \(T= \frac{\frac{ϕ ZNp}{60A}I_a}{\frac{{{N_s} \times 2\pi }}{{60}}}=\frac{ϕ ZpI_a}{2\pi A}\)

**Calculation:**

Given,

Armature current I_{a} = 50 A

Number of poles p = 4

Lap wound A = p = 4

Number of conductors Z = 480

The flux per pole ϕ = 20 mWb

\(T=\frac{ϕ ZpI_a}{2\pi A}=\frac{20\times 10^{-3}\times 480\times 50 \times4}{4\times 2\pi }\)

**T = 76.4 N-m**