1

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

A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to : A
1.0 $\times$ 10$-$7 T
B
1.5 $\times$ 10$-$7 T
C
1.5 $\times$ 10$-$5 T
D
1.0 $\times$ 10$-$5 T

## Explanation

Magnetic field due to circular arc

$\overrightarrow B = {{{\mu _0}\,i\,\theta } \over {4\pi r}}$

Due to QR arc magnetic field is outward direction of the plane.

And due to PS arc magnetic field is inward direction of the plane,

So, net magnetic field,

$\overrightarrow B = \left( {{{\overrightarrow B }_{QR}} - {{\overrightarrow B }_{PS}}} \right)\widehat K$

$\overrightarrow B = {{{\mu _0}i\theta } \over {4\pi }}\left( {{1 \over {3 \times {{10}^{ - 2}}}} - {1 \over {5 \times {{10}^{ - 2}}}}} \right)\widehat K$

$\overrightarrow B = {10^{ - 7}} \times 10 \times {\pi \over 4}\left( {{1 \over {3 \times {{10}^{ - 2}}}} - {1 \over {5 \times {{10}^{ - 2}}}}} \right)\widehat K$

$\left| {\overrightarrow B } \right| = 1 \times {10^{ - 5}}\,\,T$
2

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

A conducting circular loop made of a thin wire, has area 3.5 $\times$ 10$-$3 m2 and resistance 10 $\Omega$. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T)sin(50$\pi$t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to :
A
0.14 mC
B
0.7 mC
C
0.21 mC
D
0.6 mC

## Explanation

At    t  =  0 s

B(0) = 0.4 sin (0) = 0

and at t  =  10 ms

B(10) = 0.4 sin (50$\pi$$\times$10$\times$10-3)

= 0.4 sin $\left( {{\pi \over 2}} \right)$

= 0.4

As q = ${{\Delta \phi } \over R}$

= ${{A\left[ {B\left( {10} \right) - B\left( 0 \right)} \right]} \over {10}}$

= ${{3.5 \times {{10}^{ - 3}}\left[ {0.4 - 0} \right]} \over {10}}$

= 0.14 mC
3

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

An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d (d > > a). If the loop a applies a force F on the wire then : A
$F = 0$
B
$F \propto \left( {{a \over d}} \right)$
C
$F \propto \left( {{{{a^2}} \over {{d^3}}}} \right)$
D
$F \propto {\left( {{a \over d}} \right)^2}$

## Explanation

Magnetic field due to current carrying loop at distance d, is,

B = ${{{\mu _0}} \over {4\pi }} \times {{\overrightarrow M \times \overrightarrow d } \over {{d^3}}}$

= ${{{\mu _0}} \over {4\pi }} \times {{i \times \pi {a^2} \times d \times \sin {{90}^o}} \over {{d^3}}}$   [as  M = iA]

= ${{{\mu _0}} \over {4\pi }} \times {{i\pi {a^2}} \over {{d^2}}}$

$\therefore$   B $\propto$ ${{{a^2}} \over {{d^2}}}$

We also know,

F = Bil

$\therefore$   F $\propto$ B

So, F $\propto$ ${{{a^2}} \over {{d^2}}}$
4

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

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $\overrightarrow E = 63\widehat j\,V/m.$ The corresponding magnetic field $\overrightarrow {B,}$ at that point will be :
A
18.9 $\times$ 10$-$8 $\widehat k$T
B
2.1 $\times$ 10$-$8 $\widehat k$T
C
6.3 $\times$ 10$-$8 $\widehat k$T
D
18.9 $\times$ 108 $\widehat k$T

## Explanation

Given, $\overrightarrow E = 6.3\widehat j$

$\therefore$   $\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3$

$\therefore$   $\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}$

$= {{6.3} \over {3 \times {{10}^8}}}$

$= 2.1 \times {10^{ - 8}}\,$ T

Now we have to find the direction of $\overrightarrow B$.

We know, $\widehat E \times \widehat B = \widehat C$ and given $\overrightarrow E$ is in y-direction and wave moving in positive x-direction.

$\therefore$   $\widehat J \times \widehat B = \widehat i$

$\Rightarrow$   $\widehat B = \widehat K$

$\therefore$   $\overrightarrow B = \left| {\overrightarrow B } \right|\widehat B$

$= 2.1 \times {10^{ - 8}}\,\widehat K\,T$