Question

QUESTION is in attachment​

Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

GIVEN:-

cross sectional area = 0.05 m²

number of turns = 800

magnetic field through perpendicular to the coil B = 4 × 10^-5 Wb/m²

$\sf\xi{in}=-\frac{800\times4\times10^{-5}\times0.05\times(cos90^{\circ}-cos0^{\circ})}{0.1}$

$\sf{ \frac{(800 × 4 × 10^{-5} × 0.05 )}{0.1}}$

$\bf{160 \times {10}^{ - 4}}$

= 0.016 V

Answer 2

SOLUTION

Number of turns, n= 800

magnetic field, B= 4× 10^-5 Wb/m^2

Area, A= 0.05m^2

Initial flux= nBA

On rotating The coil by 90°, flux becomes zero.

Therefore,

e= dtheta/dt

=) e= nBA/dt

=)e= 800×4×10^-5× 0.05/0.1

=) e= 0.016 V

OPTION (d)✓

hope it helps ✔️