1

### JEE Main 2019 (Online) 10th January Morning Slot

An insulating thin rod of length $l$ has a linear charge density $\rho \left( x \right)$ = ${\rho _0}{x \over l}$ on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is -
A
${\pi \over 3}n\rho {l^3}$
B
${\pi \over 4}n\rho {l^3}$
C
$n\rho {l^3}$
D
$\pi n\rho {l^3}$

## Explanation

$\because$   M = NIA

dq = $\lambda$dx  &   A = $\pi$x2

$\int {dm} = \int {\left( x \right){{{\rho _0}x} \over \ell }} \,dx.\pi {x^2}$

M = ${{n{\rho _0}\pi } \over \ell }.\int\limits_0^\ell {{x^3}.dx} = {{n{\rho _0}\pi } \over \ell }.\left[ {{{{L^4}} \over 4}} \right]$

M = ${{n{\rho _0}\pi {\ell ^3}} \over 4}$ or ${\pi \over 4}n\rho {\ell ^3}$
2

### JEE Main 2019 (Online) 10th January Evening Slot

At some location on earth the horizontal component of earth’s magnetic field is 18 × 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45o angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is -
A
3.6 $\times$ 10$-$5 N
B
1.8 $\times$ 10$-$5 N
C
1.3 $\times$ 10$-$5 N
D
6.5 $\times$ 10$-$5 N

## Explanation

$\mu$Bsin45o = F${\ell \over 2}$sin45o

F = 2$\mu$B
3

### JEE Main 2019 (Online) 10th January Evening Slot

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then -
A
Th = 1.5 Tc
B
Th = Tc
C
Th = 2Tc
D
Th = 0.5 Tc

## Explanation

T = $2\pi \sqrt {{1 \over {\mu B}}}$

Th = $2\pi \sqrt {{{m{R^2}} \over {\left( {2\mu } \right)B}}}$

TC = $2\pi \sqrt {{{1/2m{R^2}} \over {\mu B}}}$
4

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

In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 $\times$ 10–19 C Mass of the electron = 9.1 $\times$ 10–31 kg]
A
7.5 $\times$ 10$-$4 m
B
7.5 $\times$ 10$-$3 m
C
7.5 m
D
7.5 $\times$ 10$-$2 m

## Explanation

$r = {{\sqrt {2mk} } \over {eB}} = {{\sqrt {2me\Delta v} } \over {eB}}$

$r = {{\sqrt {{{2m} \over e}.\Delta v} } \over B} = {{\sqrt {{{2 \times 9.1 \times {{10}^{ - 31}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {500} \right)} } \over {100 \times {{10}^{ - 3}}}}$

$r = {{\sqrt {{{9.1} \over {0.16}} \times {{10}^{ - 10}}} } \over {{{10}^{ - 1}}}} = {3 \over 4} \times {10^{ - 4}}$

$= 7.5 \times {10^{ - 4}}$