1

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

A magnet of total magnetic moment 10-2${\widehat i}$A-m2 is placed in a time varying magnetic field, B${\widehat i}$ (co where B = 1 Tesla and $\omega$ = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is -
A
0.014 J
B
0.028 J
C
0.01 J
D
0.007 J

## Explanation

Work done, W = $\left( {\Delta \overrightarrow \mu } \right).\overrightarrow B$

= 2 $\times$ 10$-$2 $\times$ 1 cos(0.125)

= 0.02 J
2

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

A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is -
A
2mV
B
12 mV
C
6 mV
D
1 mV

## Explanation

Potential difference between two faces perpendicular to x-axis will be
$\ell .\left( {\overrightarrow V \times \overrightarrow B } \right) = 12$mV
3

### 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 ${\pi \over 4}n\rho {\ell ^3}$
4

### 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