The flux of magnetic field through a closed conducing ϕ = at 2 + bt + c. (a) Writer the S.I. units of a, b and c. (b) If the magnitudes of a, b, and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2s.

Question

Philoid · Accepted Answer

Given:

flux of magnetic field ϕ = at² + bt + c

magnitudes of a, b and c= 0.20,0.40 and 0.60

(a) We know that,

Dimensions on both sides of an equation are equal and only terms with same dimensions can be added

Therefore

Dimensions of ϕ, at², bt and c are same

Thus units of a are given by

Now,

by faraday’s law of electromagnetic induction

…(i)

Where

ϵ =emf produced

ϕ =flux of magnetic field

therefore, units of is same as voltage i.e. Volt

Unit of b is given by

Using eqn.(i)

SI unit of c is given by

(b)

We know that

by faraday’s law of electromagnetic induction

Where

ϵ =emf produced

ϕ =flux of magnetic field

here ϵ is given by

Putting the values of a, b and t=2sec we have

Therefore, the induced emf at t=2sec is given by 1.2Volts

The flux of magnetic field through a closed conducing ϕ = at2 + bt + c.(a) Writer the S.I. units of a, b and c.(b) If the magnitudes of a, b, and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2s.

The flux of magnetic field through a closed conducing ϕ = at² + bt + c.
(a) Writer the S.I. units of a, b and c.

(b) If the magnitudes of a, b, and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2s.