Question

question is in attachment​

Answer 1

$<b>$

cross sectional area, A = 0.05 m²

number of turns , N = 800

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

a/c to question, change in magnetic flux is happening by rotation of coil. it is rotated through 90° in 0.1 sec .

so, use the formula,

$\xi_{in}=-\frac{NBA(cos\theta_f-cos\theta_i)}{\Delta t}$

here, $\sf\theta_f$ = 90° , $\sf\theta_i$ = 0° and ∆t = 0.1 sec

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

= (800 × 4 × 10^-5 × 0.05 )/0.1

= (8 × 4 × 0.05 × 10^-3)/0.1

= (160 × 10^-4)

= 0.016V

hence, option (d) is correct choice.

Answer 2

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

GIVEN:-

cross sectional area = 0.05 m² 

number of turns = 800 

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

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

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

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

= 0.016V 

$\huge \bf{THANKS}$