1

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

In a certain region static electric and magnetic fields exist. The magnetic field is given by $\overrightarrow B = {B_0}\left( {\widehat i + 2\widehat j - 4\widehat k} \right)$ . If a test charge moving with a velocity $\overrightarrow \upsilon = {\upsilon _0}\left( {3\widehat i - \widehat j + 2\widehat k} \right)$ experiences no force in that region, then the electric field in the region, in SI units, is :
A
$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {3\widehat i - 2\widehat j - 4\widehat k} \right)$
B
$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {\widehat i + \widehat j + 7\widehat k} \right)$
C
$\overrightarrow E = {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)$
D
$\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)$

## Explanation

Here test charge experience no net force, So, sum of electric and magnetic field is zero.

$\therefore\,\,\,$ Fe + Fm = 0

$\therefore\,\,\,$ Fe = $-$ q ($\overrightarrow v$ $\times$ $\overrightarrow B)$

= $-$ qB0 $\upsilon$0 [(3$\widehat i$ $-$ $\widehat j$ + 2$\widehat k$) $\times$ ($\widehat i$ + 2$\widehat j$ $-$ 4$\widehat k$)]

= $-$ q$\upsilon$0 B0 (14$\widehat j$ + 7$\widehat k$)

Electric field produced by the charge q,

$\overrightarrow E$ = ${{\overrightarrow {{F_e}} } \over q}$

= ${{ - q{\upsilon _0}{B_0}\left( {14\widehat j + 7\widehat k} \right)} \over q}$

= $-$ $\upsilon$0 B0 (14 ${\widehat j}$ + 7 ${\widehat k}$)
2

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

A magnetic dipole in a constant magnetic field has :
A
maximum potential energy when the torque is maximum.
B
zero potential energy when the torque is minimum.
C
zero potential energy when the torque is maximum.
D
minimum potential energy when the torque is maximum.

## Explanation

In uniform magnetic field, the torque experienced by the magnetic dipole is $\tau$ = MB sin $\theta$

Torque will be maximum when $\theta$ = 90o

$\tau$max = MB sin90o = MB

Potential energy of magnetic dipole,

$\mu$ = $-$ MB cos $\theta$

at maximum torque,

$\mu$ = $-$ MB cos 90o = 0
3

### 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
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### 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