1
AIPMT 2011 Prelims
MCQ (Single Correct Answer)
+4
-1
A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in uniform magnetic field acting along AB. If the magnetic force on the arm BC is $$\overrightarrow {F,}$$ the force on the arm AC is

A
$$- \sqrt 2 \,\overrightarrow F$$
B
$$-$$ $$\,\overrightarrow F$$
C
$$\,\overrightarrow F$$
D
$$\sqrt 2 \,\overrightarrow F$$
2
AIPMT 2010 Mains
MCQ (Single Correct Answer)
+4
-1
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is $$i$$. The resultant magnetic field due to the two semicircular parts at their common centre is
A
$${{{\mu _0}i} \over {2\sqrt 2 R}}$$
B
$${{{\mu _0}i} \over {2R}}$$
C
$${{{\mu _0}i} \over {4R}}$$
D
$${{{\mu _0}i} \over {\sqrt 2 R}}$$
3
AIPMT 2010 Mains
MCQ (Single Correct Answer)
+4
-1
A particle having a mass of 10$$-$$2 kg carries a charge of 5 $$\times$$ 10$$-$$8 C. The particle is given an initial horizontal velocity of 105 m s$$-$$1 in the presence of electric field $$\overrightarrow E$$ and magnetic field $$\overrightarrow B$$. To keep the particle moving in a horizontal direction, it is necessary that
(1)  $$\overrightarrow B$$ should be perpendicular to the direction of velocity and $$\overrightarrow E$$ should be along the direction of velocity

(2)  Both $$\overrightarrow B$$ and $$\overrightarrow E$$ should be along the direction of velocity

(3)  Both $$\overrightarrow B$$ and $$\overrightarrow E$$ are mutually perpendicular and perpendicular to the direction of velocity.

(4)  $$\overrightarrow B$$ should be along the direction of velocity and $$\overrightarrow E$$ should be perpendicular to the direction of velocity

Which one of the following pairs of statements is possible ?
A
(1) and (3)
B
(3) and (4)
C
(2) and (3)
D
(2) and (4)
4
AIPMT 2010 Mains
MCQ (Single Correct Answer)
+4
-1
A closely wound solenoid of 2000 turns and area of cross-section 1.5 $$\times$$ 10$$-$$4 m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 $$\times$$ 10$$-$$2 tesla making an angle of 30o with the axis of the solenoid. The torque on the solenoid will be
A
3 $$\times$$ 10$$-$$3 N m
B
1.5 $$\times$$ 10$$-$$3 N m
C
1.5 $$\times$$ 10$$-$$2 N m
D
3 $$\times$$ 10$$-$$2 N m
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