1

### JEE Main 2019 (Online) 9th January Evening Slot

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the radio of the magnetic field at the central of the loop (BL) to that at the center of the coil (BC), i.e. ${{{B_L}} \over {{B_C}}}$ will be :
A
N
B
${1 \over N}$
C
N2
D
${1 \over {{N^2}}}$

## Explanation

For loop,

L = 2$\pi$R

For coil,

L = N $\times$ 2$\pi$r

$\therefore$   2$\pi$R = N $\times$ 2$\pi$r

$\Rightarrow$  R = Nr

$\Rightarrow$  r = ${R \over N}$

We know,

BL = ${{{\mu _0}i} \over {2R}}$

and  BC = N $\times$ ${{{\mu _0}i} \over {2r}}$

$\therefore$  ${{{B_L}} \over {{B_C}}} = {{{{{\mu _0}i} \over {2R}}} \over {N \times {{{\mu _0}i} \over {2\left( {{R \over N}} \right)}}}} = {1 \over {{N^2}}}$
2

### 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
3

### 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
4

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