1

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

A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 m, 100 turns, and carrying a current of 5.2 A. The corecivity of the bar magnet is :
A
285 A/m
B
2600 A/m
C
520 A/m
D
1200 A/m

## Explanation

Coercivity, H = ${B \over {{\mu _0}}}$

Inside solenoid the magnetic field,

B = $\mu$0ni

$\therefore$   H = ${{{\mu _0}ni} \over {{\mu _0}}}$

= ni

= ${N \over \ell } \times i$

= ${{100} \over {0.2}} \times 5.2$

= 2600 A/m
2

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

The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :
A
${U_E} = {{{U_B}} \over 2}$
B
${U_E} > {U_B}$
C
${U_E} < {U_B}$
D
${U_E} = {U_B}$

## Explanation

Energy density of magnetic field (UB) = ${{{B^2}} \over {2{\mu _0}}}$

Also,

Energy density of electric field

UE = ${1 \over 2}$ $\varepsilon$0E2

= ${1 \over 2}$ $\varepsilon$0 B2 C2     [as   ${E \over B}$ = C ]

= ${1 \over 2}$ $\varepsilon$0 B2 $\times$ $\left( {{1 \over {{\varepsilon _0}{\mu _0}}}} \right)$    [as   C2 = ${{1 \over {{\varepsilon _0}{\mu _0}}}}$]

= ${{{B^2}} \over {2{\mu _0}}}$

$\therefore$  UB = UE
3

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

A particle having the same charge as of electron moves in a ciurcular path of radius 0.5 cm under the influence of a magnetic field 0f 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 $\times$ 10$-$19C)
A
9.1 $\times$ 10$-$31 kg
B
1.6 $\times$ 10$-$27 kg
C
1.6 $\times$ 10$-$19 kg
D
2.0 $\times$ 10$-$24 kg

## Explanation

Given,
radius of circular path(r) = 0.5 cm
Magnetic field (B) = 0.5 T
Electric field (E) = 100 V/m

Charge of particle (q) = 1.6$\times$10$-$19 C

As particle is moving in a circular path so,

${{m{v^2}} \over r} = qvB$

$\Rightarrow$  r = ${{mv} \over {qB}}$ . . . . . . . (1)

When electric field of 100 v/m is applied on the particle then particle is moving in the straight line. So, the net force on the particle is zero.

$\therefore$    Fnet = 0

$\Rightarrow$   Fe = Fm

$\Rightarrow$  qE = qvB

$\Rightarrow$   E = vB  . . . . .(2)

From equation (1) and (2) we get,

r = ${m \over {qB}}\left( {{E \over B}} \right)$

= ${{mE} \over {q{B^2}}}$

$\Rightarrow$   m = ${{q{B^2}r} \over E}$

= ${{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}$

= 2 $\times$ 10$-$24 kg
4

### 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 $\times$ ${{{\mu _0}i} \over {2r}}$

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