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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

MCQ (Single Correct Answer)
A charge $$Q$$ is uniformly distributed over the surface of non-conducting disc of radius $$R.$$ The disc rotates about an axis perpendicular to its plane and passing through its center with an angular velocity $$\omega .$$ As a result of this rotation a magnetic field of induction $$B$$ is obtained at the center of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and very the radius of the disc then the variation of the magnetic induction at the center of the disc will be represented by the figure :
A
B
C
D

Explanation

The magnetic field due a disc is given as

$$B = {{{h_0}\omega Q} \over {2\pi R}}$$ i.e., $$B \propto {1 \over R}$$
2

AIEEE 2012

MCQ (Single Correct Answer)
Proton, deuteron and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, denuteron and alpha particle are respectively $${r_p},{r_d}$$ and $${r_\alpha }$$. Which one of the following relation is correct?
A
$${r_\alpha } = {r_p} = {r_d}$$
B
$${r_\alpha } = {r_p} < {r_d}$$
C
$${r_\alpha } > {r_d} > {r_p}$$
D
$${r_\alpha } = {r_d} > {r_p}$$

Explanation

$$r = {{\sqrt {2mv} } \over {qB}} \Rightarrow r \times v{{\sqrt m } \over q}$$

Thus we have, $${r_\alpha } = {r_p} < {r_d}$$
3

AIEEE 2011

MCQ (Single Correct Answer)
A current $$I$$ flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius $$R.$$ The magnitude of the magnetic induction along its axis is:
A
$${{{\mu _0}I} \over {2{\pi ^2}R}}$$
B
$${{{\mu _0}I} \over {2\pi R}}$$
C
$${{{\mu _0}I} \over {4\pi R}}$$
D
$${{{\mu _0}I} \over {{\pi ^2}R}}$$

Explanation

Current in a small element, $$dl = {{d\theta } \over \pi }I$$

Magnetic field due to the element

$$dB = {{{\mu _0}} \over {4\pi }}{{2dl} \over R}$$

The component $$dB$$ $$\cos \,\theta ,$$ of the field is canceled by another opposite component.

Therefore,



$${B_{net}} = \int {dB\sin \theta = {{{\mu _0}I} \over {2{\pi ^2}{R_0}}}} $$

$$\int\limits_0^\pi {\sin \theta d\theta = {{{\mu _0}I} \over {{\pi ^2}R}}} $$
4

AIEEE 2010

MCQ (Single Correct Answer)
Two long parallel wires are at a distance $$2d$$ apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field $$B$$ along the line $$XX'$$ is given by
A
B
C
D

Explanation

The magnetic field varies inversely with the distance for a long conductor. That is, $$B \propto {1 \over d}.$$ According to the magnitude and direction shown graph $$(1)$$ is the correct one.

Questions Asked from Magnetics

On those following papers in MCQ (Single Correct Answer)
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