1

### JEE Main 2017 (Online) 8th April Morning Slot

Magnetic field in a plane electromagnetic wave is given by

$\overrightarrow B$ = B0 sin (k x + $\omega$t) $\widehat j\,T$

Expression for corresponding electric field will be :
Where c is speed of light.
A
$\overrightarrow E$ = B0 c sin (k x + $\omega$t) $\widehat k$ V/m
B
$\overrightarrow E$ = ${{{B_0}} \over c}$ sin (k x + $\omega$t) $\widehat k$ V/m
C
$\overrightarrow E$ = $-$ B0 c sin (kx +$\omega$t) $\widehat k$ V/m
D
$\overrightarrow E$ = B0 c sin (kx $-$$\omega t) \widehat k V/m ## Explanation The relation between electric and magnetic field is , C = {{\overrightarrow E } \over {\overrightarrow B }} \Rightarrow$$\,\,\,$ $\overrightarrow E$ = C $\overrightarrow B$

Electric field component is perpendicular to the direction of magnetic field. Given magnetic field is along y $-$ axis,

So, electric field along z $-$ axis will be

$\overrightarrow E$ = B0 C sin (kx + $\omega$t) $\,\widehat k$ v/m
2

### JEE Main 2017 (Online) 9th April Morning Slot

A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is :
A
away from the wire
B
towards the wire
C
parallel to the wire along the current
D
parallel to the wire opposite to the current
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### JEE Main 2017 (Online) 9th April Morning Slot

A uniform magnetic field B of 0.3 T is along the positive Z-direction. A rectangular loop (abcd) of sides 10 cm × 5 cm carries a current I of 12 A. Out of the followingdifferent orientations which one corresponds to stable equilibrium ?
A B C D 4

### JEE Main 2018 (Offline)

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, r$_\alpha$ respectively in a uniform magnetic field B. The relation between re, rp, r$_\alpha$ is:
A
re < r$_\alpha$ < rp
B
re > rp = r$_\alpha$
C
re < rp = r$_\alpha$
D
re < rp < r$_\alpha$

## Explanation

When a charged particle moves in a magnetic field then the charged particle moves in a circular path. So,

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

$\Rightarrow $$\,\,\, r = {{mv} \over {Bq}} We know kinetic energy, K = {1 \over 2} mv2 \therefore\,\,\, mv = \sqrt {2Km} \therefore\,\,\, r = {{\sqrt {2Km} } \over {Bq}} According to the question, Ke (electron) = Kp (proton) = K\alpha (Alpha particle) = K = constant, and all of them are in uniform magnetic field. \therefore B = constant. \therefore\,\,\, r \propto {{\sqrt m } \over q} For proton (1H1), mass = m, and charge = e \therefore\,\,\, rp \propto {{\sqrt m } \over e} For alpha particle (2H4), mass = 4m and charge = 2e \therefore\,\,\, r \alpha \propto {{\sqrt {4m} } \over {2e}} \propto {{\sqrt m } \over e} \therefore$$\,\,\,$ rp = r$\alpha$

For electron,

charge = e

and mass (me) = 9.1 $\times$ 10$-$31 kg

and mass of proton = 1.67 $\times$ 10$-$27 kg

$\therefore\,\,\,$ mass of electron < mass of proton.

re $\propto$ ${{\sqrt {{m_e}} } \over e}$ < rp

$\therefore\,\,\,$ re < rp = r$\propto$