1

### JEE Main 2018 (Online) 16th April Morning Slot

A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity $\omega ,$ the maxium e.m.f. induced in the coil will be:
A
3 nBA$\omega$
B
${3 \over 2}$ nBA$\omega$
C
nBA$\omega$
D
${1 \over 2}$ nBA$\omega$

## Explanation

Flux in the coil, $\phi$ = nBA sin($\omega$t)

When n = no. of turns

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ A = Area of coil

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$\omega$ = angular speed

Induced emf,

$\left| e \right| = {{d\phi } \over {dt}}$

= nBA$\omega$ cos$\omega$t

$\therefore\,\,\,$ emax = nBA$\omega$
2

### JEE Main 2018 (Online) 16th April Morning Slot

A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ;
A
50 A
B
45 A
C
25 A
D
20 A

## Explanation

Efficiency n = 0.9 = ${{{P_s}} \over {{P_p}}}$

as $\,\,\,\,\,\,$ P = VI

$\therefore\,\,\,$ Ps = 0.9 $\times$ Pp

$\Rightarrow$ $\,\,\,\,$ Vs Is = 0.9 $\times$ Vp Ip

$\Rightarrow$ $\,\,\,\,$ Is = ${{0.9 \times 2300 \times 5} \over {230}}$

= $\,\,\,\,$ 45 A
3

### JEE Main 2018 (Online) 16th April Morning Slot

A plane electromagnetic wave of wavelength $\lambda$ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by :
A
B
C
D
4

### JEE Main 2019 (Online) 9th January Evening Slot

A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :
A
50 A
B
45 A
C
35 A
D
25 A

## Explanation

Given,

Primary voltage (VP) = 2300 V

Primary current (IP) = 5A

Secondary voltage (VS) = 230 V

efficiency ($\eta$) = 90%

We know,

Efficiency ($\eta$) = ${{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}$

$\therefore$   $\eta$ = ${{{P_S}} \over {{P_P}}}$ = ${{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}$

$\Rightarrow$   0.9 = ${{230 \times {{\rm I}_S}} \over {2300 \times 5}}$

$\Rightarrow$   I = 45 A

NEET