Question

1. There are 20 turns in the armature of a coil having 0.127m^2 area. How fast it should be rotated in a uniform
magnetic field of 0.2 Wb/m2 so that emf induced is 160
volt?​

Answer 1

Explanation:

the answer is 50 rps

What is the name of the small boy?One who treats the condition of teeth and oral cavity is called _____. During corona lock down, a girl carries her father to the farthest hospital. In

that time, which muscle functions more?

a) Striated Muscle voluntary Muscle

b) Striated Muscle, involuntary muscle

c) Non Striated Muscle, Voluntary muscle

d) Non Striated Muscle, involuntary muscle

Answer 2

The armature of the coil should be rotated 315 rad/sec or 50 rps so that the peak value of e.m.f induced in it is 160 volt.

Given:

E(t) = 160V

B = 0.2 Wb/m²

A = 0.127m²

N = 20 turns

To Find:

The frequency for rotating the armature of a coil so that induced emf would be 160V.

Solution:

We know that,

E = E₀sin(wt) where, E₀ = BAWN

In peak the value of sin(wt) = 1. Therefore, to find the angular frequency the values of B,A,N,and E are substituted which is given by,

160 = 0.2 x 0.127 x 20 x N x 1

160 = 0.508 N

N = 160/0.508

N = 314.9606 rad/s

N = 315 rad/s (approx)

We know that,

2π/60 = 1rps

1 rad/s = 0.159 rps

Therefore, for 315 rad/s the rps is given by

=315x0.159 rps

= 50 rps

Therefore, the armature of the coil should rotate 315rad/s or 50 rps so that induce emf of 160V could be produced.

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