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### 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
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### 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 ${\pi \over 4}n\rho {\ell ^3}$
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### 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
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### 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 = $2\pi \sqrt {{{1/2m{R^2}} \over {\mu B}}}$