There is part A,B, please show work and answers thanks. An electromagnetic doorb

Question

There is part A,B, please show work and answers thanks.

An electromagnetic doorbell has been constructed by wrapping a wire into a 70 turn coil around a long thin rod as shown in the figure. The rod has a mass of 30 g, a length of 8 cm, and a cross-sectional area of 0.2 cm2. The rod is free to pivot about an axis through its center, which is also the center of the coil. Initially, the rod makes an angle of 9 = 25 degree . When 0 = 0, the rod strikes a bell. There is a uniform magnetic field of 0.09 T directed along 0 = 0. If a current of 2 A is run through the coil, what is the torque on the rod when = 25 degree ? What is the angular velocity of the rod when it strikes the bell? To do the calculation, you need to know the moment of inertia of the thin rod. The table below will help. Please note that the torque is dynamically changing

Explanation / Answer

a. torque acting on a body T = NiAB sin theta

T = 70 *2 * 0.2*10^-4 * 0.09 * sin 25

T = 4.22*10^-5 N-m

b. now torque T = I alpha

alpha = T/I = 4.22*10^-5/(ML^2/12)

ALPHA = 4.22*10^-5/(0.03*0.08^2/12)

ALPHA = 2.63 rad/s^2

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