Question

A circular coil (950 turns, radius = 0.060 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.010 s, the normal makes an angle of ϕ = 45° with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.065 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

Answer 1

B = 8.55 × 10⁻⁵ T

Step-by-step explanation:

Data Provided:

Number of turns, N = 950 turns,

Radius of coil, r = 0.060 m

Now,

Magnetic flux (φ) is given as:

φ = BAcosθ

where,

B is the magnetic field

A is the area

Now,

At t = 0 s, the normal to the coil is perpendicular to the magnetic field, i.e θ = 90°

Thus,

φ = BAcos90° = 0.

Also,

At t = 0.010 s, the normal makes an angle θ = 45°

Thus,

φ= NBAcos(45°),

A = π × r²

Average emf induced in the coil, E = 0.065 V .

Thus,

E = NBAcos(45°)/t

or

E = NB×πr²×cos(45°)/t

or

B = (E × t)/(NB×πr²×cos45°)

on substituting the values

we get,

B = (0.065 V × 0.010 s) / (950 × π(0.06 )² × cos 45°)

or

B = 8.55 × 10⁻⁵ T