A rectangular coil of two turns having area A rotates in a uniform magnetic field B with angular speed ω about an axis perpendicular to the field and in plane of the coil. If initially the plane of the coil is perpendicular to the field, then the average induced emf when it has rotated through (π/2) is

Question

A rectangular coil of two turns having area A rotates in a uniform magnetic field B with angular speed ω about an axis perpendicular to the field and in plane of the coil. If initially the plane of the coil is perpendicular to the field, then the average induced emf when it has rotated through (π/2) is (A)  (4BAω/π) (B) 2BAω (C)  (2BAω/π) (D)  (BAω/2π)

Tardigrade Admin · Accepted Answer

ε = NBAω sinω t ε = 2 BAω sinω t εavg = (2BAω∫(t/4)0sinω t dt /(T)4) = (8BAω/2π) ([-cos ω t]o(T/4)/ω) = (8BAω/ω T).[1-0]=(8BAω/2π) ϵavg = (4BAω/π)

$\frac{4 B A ω}{π}$

$2 B A ω$

$\frac{2 B A ω}{π}$

$\frac{B A ω}{2 π}$

$ε = NB A ω s inω t$
$ε = 2 B A ω s inω t$
$ε_{a vg} = \frac{2 B A ω \int _{0}^{\frac{t}{4}} s inω t d t}{\frac{T}{4}} = \frac{8 B A ω}{2 π} \frac{[ - cos ω t ] _{o^{\frac{T}{4}}}}{ω}$
$= \frac{8 B A ω}{ω T} . [1 - 0] = \frac{8 B A ω}{2 π}$
$ϵ_{a vg} = \frac{4 B A ω}{π}$

π4BAω​

2BAω

π2BAω​

2πBAω​

ε=NBAωsinωt ε=2BAωsinωt εavg​=4T​2BAω∫04t​​sinωtdt​=2π8BAω​ω[−cosωt]o4T​​​ =ωT8BAω​.[1−0]=2π8BAω​ ϵavg​=π4BAω​

$\frac{4 B A ω}{π}$

$2 B A ω$

$\frac{2 B A ω}{π}$

$\frac{B A ω}{2 π}$

$ε = NB A ω s inω t$
$ε = 2 B A ω s inω t$
$ε_{a vg} = \frac{2 B A ω \int _{0}^{\frac{t}{4}} s inω t d t}{\frac{T}{4}} = \frac{8 B A ω}{2 π} \frac{[ - cos ω t ] _{o^{\frac{T}{4}}}}{ω}$
$= \frac{8 B A ω}{ω T} . [1 - 0] = \frac{8 B A ω}{2 π}$
$ϵ_{a vg} = \frac{4 B A ω}{π}$