1

### JEE Main 2019 (Online) 11th January Evening Slot

A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by :
[Given permittivity of space $\in$0 = 9 $\times$ 10–12 SI units, Speed of light c = 3 $\times$ 108 m/s]
A
2 kV/m
B
1 kV/m
C
1.4 kV/m
D
0.7 kV/m

## Explanation

Intensity of EM wave is given by

${\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C$

$= {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$ $= {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}$

$E = \sqrt 2 \times {10^3}kv/m$

$= 1.4$ kv/m
2

### JEE Main 2019 (Online) 11th January Evening Slot

A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 $\times$ 10–6 J/ T when a magnetic intensity of 60 $\times$ 103 A/m is applied. Its magnetic susceptibility is
A
3.3 $\times$ 10–4
B
2.3 $\times$ 10–2
C
4.3 $\times$ 10–2
D
3.3 $\times$ 10–2

## Explanation

x = ${1 \over H}$

I = ${{Magnetic\,moment} \over {Volume}}$

I = ${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$ = 20 N/m2

x = ${{20} \over {60 \times {{10}^{ + 3}}}}$ = ${1 \over 3} \times {10^{ - 3}}$

= 0.33 $\times$ 10$-$3 = 3.3 $\times$ 10$-$4
3

### JEE Main 2019 (Online) 12th January Evening Slot

A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms–1, at right angles to the horizontal component of the earth's magnetic field of 0.3 $\times$ 10–4 Wb/m2. The value of the induced emf in wire is :
A
0.3 $\times$ 10–3 V
B
2.5 $\times$ 10–3 V
C
1.5 $\times$ 10–3 V
D
1.1 $\times$ 10–3 V

## Explanation

Induied emf = Bv$\ell$ sin 45o

= 0.3 $\times$ 10$-$4 $\times$ 5 $\times$ 10 $\times$ sin 45o

= 1.1 $\times$ 10$-$3 V
4

### JEE Main 2019 (Online) 12th January Evening Slot

A paramagnetic material has 1028 atoms/m3. Its magnetic susceptibility at temperature 350 K is 2.8 $\times$ 10–4. Its susceptibility at 300 K is :
A
3.726 $\times$ 10–4
B
2.672 $\times$ 10–4
C
3.672 $\times$ 10–4
D
3.267 $\times$ 10$-$4

## Explanation

x $\alpha$ ${1 \over {{T_C}}}$

curie law for paramagnetic substane

${{{x_1}} \over {{x_2}}}$ = ${{{T_{{C_2}}}} \over {{T_{{C_1}}}}}$

${{2.8 \times {{10}^{ - 4}}} \over {{x_2}}} = {{300} \over {350}}$

x2 = ${{2.8 \times 350 \times {{10}^{ - 4}}} \over {300}}$

= 3.266 $\times$ 10$-$4