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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 2

### 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$\propto$
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### JEE Main 2018 (Offline)

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is ${{B_2}}$. The ratio ${{{B_1}} \over {{B_2}}}$ is:
A
2
B
$\sqrt 3$
C
$\sqrt 2$
D
$1 \over \sqrt 2$

## Explanation

Dipole moment, M = IA

Let radius of circular loop = R

$\therefore\,\,\,$ M  =  I $\times$ $\pi$R2

Later, we keep current constant ,

But dipole moment becomes double, let new radius = R1

$\therefore\,\,\,$ 2M  =  I $\times$ $\pi$R$_1^2$

$\Rightarrow $$\,\,\, 2I\pi R2 = I\pi R_1^2 \Rightarrow$$\,\,\,$ R1  =  $\sqrt 2$ R

At the center of circular ring, the magnetic field,

B  =  ${{{\mu _0}I} \over {2R}}$

$\therefore\,\,\,$ B1  =  ${{{\mu _0}I} \over {2R}}$   and    B2 = ${{{\mu _0}I} \over {2 \times \left( {\sqrt 2 R} \right)}}$

$\therefore\,\,\,$ ${{{B_1}} \over {{B_2}}}$  =   ${{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}$

= ${{2\sqrt 2 } \over 2}$

= $\sqrt 2$
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### JEE Main 2018 (Online) 15th April Morning Slot

A Helmholtz coil has a pair of loops, each with $N$ turns and radius $R$. They are placed coaxially at distance $R$ and the same current ${\rm I}$ flows through the loops in the same direction. $P,$ midway between the centers $A$ and $C$, is given by [Refer to figure given below] : A
${{8N{\mu _0}{\rm I}} \over {{5^{1/2}}R}}$
B
${{8N{\mu _0}{\rm I}} \over {{5^{3/2}}R}}$
C
${{4N{\mu _0}{\rm I}} \over {{5^{1/2}}R}}$
D
${{4N{\mu _0}{\rm I}} \over {{5^{3/2}}R}}$

## Explanation

P is the mid-point of line AC. A and C are the center of the two circle of each radius R.

Current flows through loop A and B are in same direction, So the magnetic field will also be in the same direction. Magnitude of magnetic field at paint P

= magnitude of magnetic field due to A and B at paint P.

= 2 $\left[ {{{{\mu _0}NI\,{R^2}} \over {2{{\left( {{R^2} + {{{R^2}} \over 4}} \right)}^{{3 \over 2}}}}}} \right]$

$=$ ${{{\mu _0}\,NI{R^2}} \over {{{{5^{{3 \over 2}}}{R^3}} \over 8}}}$

= ${{8\,{\mu _0}NI} \over {{5^{{3 \over 2}}}\,R}}$