1

### 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
2

### 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
3

### 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
4

### JEE Main 2019 (Online) 9th April Morning Slot

The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to :
A
1/L2
B
1/L
C
L
D
L2

## Explanation

$\phi$ = NBA = LI

N $\mu$0 nI$\pi$R2 = LI

$N{\mu _0}{N \over l}l\pi {R^2} = LI$

N and R constant

Self inductance (L) $\propto {1 \over l} \propto {1 \over {length}}$