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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Two long current carrying thin wires, both with current $$I,$$ are held by insulating threads of length $$L$$ and are in equilibrium as shown in the figure, with threads making an angle $$'\theta '$$ with the vertical. If wires have mass $$\lambda $$ per unit-length then the value of $$I$$ is :
($$g=$$ $$gravitational$$ $$acceleration$$ )
A
$$2\sqrt {{{\pi gL} \over {{\mu _0}}}\tan \theta } $$
B
$$\sqrt {{{\pi \lambda gL} \over {{\mu _0}}}\tan \theta } $$
C
$$\sin \theta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos \theta }}} $$
D
$$2\sin \theta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos \theta }}} $$

Explanation

Let us consider $$'\ell '$$ length of current carrying wire,

At equilibrium

$$T\cos \theta = \lambda g\ell $$



and $$T\sin \theta = {{{\mu _0}} \over {2\pi }}{{I \times Il} \over {2L\sin \theta }}$$

$$\left[ {\,\,} \right.$$ as $$\left. {{{{F_B}} \over \ell } = {{{\mu _0}} \over {4\pi }}{{2I \times I} \over {2\ell \sin \theta }}\,\,} \right]$$

Therefore, $$I = 2\sin \theta \sqrt {{{\pi \lambda gL} \over {{u_0}\cos \theta }}} $$
2

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
A conductor lies along the $$z$$-axis at $$ - 1.5 \le z < 1.5\,m$$ and carries a fixed current of $$10.0$$ $$A$$ in $$ - {\widehat a_z}$$ direction (see figure). For a field $$\overrightarrow B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}\,\,{\widehat a_y}\,\,T,$$ find the power required to move the conductor at constant speed to $$x=2.0$$ $$m$$, $$y=0$$ $$m$$ in $$5 \times {10^{ - 3}}s.$$ Assume parallel motion along the $$x$$-axis.
A
$$1.57W$$
B
$$2.97W$$
C
$$14.85$$ $$W$$
D
$$29.7W$$

Explanation

Work done in moving the conductor is,

$$W = \int_0^2 {Fdx} $$

$$ = \int_0^2 {3.0 \times {{10}^{ - 4}}\,{e^{ - 02x}} \times 10 \times 3dx} $$



$$ = 9 \times {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}\,dx} $$

$$ = {{9 \times {{10}^{ - 3}}} \over {0.2}}\left[ { - {e^{ - 0.2 \times 2}} + 1} \right]\,\,\,$$

$$B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}$$

(By exponential function)

$$ = {{9 \times {{10}^{ - 3}}} \over {0.2}} \times \left[ {1 - {e^{ - 0.4}}} \right]$$

$$ = 9 \times {10^{ - 3}} \times \left( {0.33} \right) = 2.97 \times {10^{ - 3}}J$$

Power required to move the conductor is,

$$P = {W \over t}$$

$$P = {{2.97 \times {{10}^{ - 3}}} \over {\left( {0.2} \right) \times 5 \times {{10}^{ - 3}}}} = 2.97W$$
3

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
The coercivity of a small magnet where the ferromagnet gets demagnetized is $$3 \times {10^3}\,A{m^{ - 1}}.$$ The current required to be passed in a solenoid of length $$10$$ $$cm$$ and number of turns $$100,$$ so that the magnet gets demagnetized when inside the solenoid, is :
A
$$30$$ $$mA$$
B
$$60$$ $$mA$$
C
$$3$$ $$A$$
D
$$6A$$

Explanation

Magnetic field in solenoid $$B = {\mu _0}ni$$

$$ \Rightarrow {B \over {{\mu _0}}} = ni$$

(Where $$n=$$ number of turns per unit length)

$$ \Rightarrow {B \over {{\mu _0}}} = {{Ni} \over L}$$

$$ \Rightarrow 3 \times {10^3} = {{100i} \over {10 \times {{10}^{ - 2}}}}$$

$$ \Rightarrow i = 3A$$
4

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
Two short bar magnets of length $$1$$ $$cm$$ each have magnetic moments $$1.20$$ $$A{m^2}$$ and $$1.00$$ $$A{m^2}$$ respectively. They are placed on a horizontal table parallel to each other with their $$N$$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $$20.0$$ $$cm.$$ The value of the resultant horizontal magnetic induction at the mid-point $$O$$ of the line joining their centres is close to $$\left( \, \right.$$ Horizontal com-ponent of earth. $$s$$ magnetic induction is $$3.6 \times 10.5Wb/{m^2})$$
A
$$3.6 \times 10.5\,\,Wb/{m^2}$$
B
$$2.56 \times 10.4\,\,Wb/{m^2}$$
C
$$3.50 \times 10.4\,\,Wb/{m^2}$$
D
$$5.80 \times 10.4\,Wb/{m^2}$$

Explanation



Given: $${M_1} = 1.20A{m^2}\,\,\,$$ and $$\,\,\,{M_2} = 1.00A{m^2}$$

$$r = {{20} \over 2}cm = 0.1m$$

$${B_{net}} = {B_1} + {B_2} + {B_H}$$

$${B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}$$

$$ = {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}$$

$$ = 2.56 \times {10^{ - 4}}\,\,wb/{m^2}$$

Questions Asked from Magnetics

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