Practice Start Test

Magnetic Effect of Current

NEET Physics · 96 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
A square loop, carrying a teady current II, is placed in a horizontal plane near a long straight conductor carrying a steady current II 1 at a distance d from the conductor as shown in figure. The loop will experience
A a net attractive force towards the conductor
B a net repulsive force away from the conductor
C a net torque acting upward perpendicular to the horizontal plane
D a net torque acting downward normal to horizontal plane
Correct Answer
Option A
Solution
F1>F2{F_1} > {F_2}

as

F∝1dF \propto {1 \over d}

, and

F3{F_3}

and

F4{F_4}

are equal and opposite. Hence, the net attraction force will be towards the conductor.

Q42
Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency ff Hz. The magnitude of magnetic induction at the center of the ring is
A μ0qf2πR{{{\mu _0}qf} \over {2\pi R}}
B μ0qf2R{{{\mu _0}qf} \over {2R}}
C μ0q2fR{{{\mu _0}q} \over {2fR}}
D μ0q2πfR{{{\mu _0}q} \over {2\pi fR}}
Correct Answer
Option B
Solution

When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is I =

qT{q \over T}

= qf Magnetic field at the centre of the ring is B =

μ0I2R{{{\mu _0}I} \over {2R}}

=

μ0qf2R{{{\mu _0}qf} \over {2R}}
Q43
A galvanometer of resistance, G, is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
A G(S+G){G \over {(S + G)}}
B S2(S+G){{{S^2}} \over {\left( {S + G} \right)}}
C (SGS+G)\left( {{{SG} \over {S + G}}} \right)
D G2(S+G){{{G^2}} \over {(S + G)}}
Correct Answer
Option D
Solution

To keep the main current in the circuit unchanged, the resistance of the galvanometer should be equal to the net resistance.

∴\therefore

G=(GSG+S)+S′G = \left( {{{GS} \over {G + S}}} \right) + S'
⇒G−GSG+S=S′\Rightarrow G - {{GS} \over {G + S}} = S'
∴S′=G2G+S\therefore S' = {{{G^2}} \over {G + S}}
Q44
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 F,→\overrightarrow {F,} the force on the arm AC is
A −2 F→ - \sqrt 2 \,\overrightarrow F
B −-  F→\,\overrightarrow F
C  F→\,\overrightarrow F
D 2 F→\sqrt 2 \,\overrightarrow F
Correct Answer
Option B
Solution

Here,

F→BC=F→{\overrightarrow F _{BC}} = \overrightarrow F
∵F→AB=0\because {\overrightarrow F _{AB}} = 0

The net magnetic force on a current carrying closed loop in a uniform magnetic field is zero.

∴F→AB+F→BC+F→AC=0\therefore {\overrightarrow F _{AB}} + {\overrightarrow F _{BC}} + {\overrightarrow F _{AC}} = 0
⇒F→AC=−F→BC\Rightarrow {\overrightarrow F _{AC}} = - {\overrightarrow F _{BC}}

(∵\because

F→AB=0{\overrightarrow F _{AB}} = 0

)

Q45
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
A will turn towards right of direction of motion
B speed will decrease
C speed will increase
D will turn towards left of direction of motion
Correct Answer
Option B
Solution
v→\overrightarrow v

and

B→\overrightarrow B

are in same direction so that magnetic force on electron becomes zero, only electric force acts.

But force on electron due to electric field is opposite to the direction of velocity.

Q46
A closely wound solenoid of 2000 turns and area of cross-section 1.5 × \times 10 −-4 m 2 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 30 o 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
Correct Answer
Option C
Solution

Magnetic moment of the loop. M = NIA = 2000 × 2 × 1.5 × 10 –4 = 0.6 J/T Torque τ\tau = MBsin30°

=0.6×5×10−2×12=1.5×10−2 Nm= 0.6 \times 5 \times {10^{ - 2}} \times {1 \over 2} = 1.5 \times {10^{ - 2}}\,Nm
Q47
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 10 5 m s −-1 in the presence of electric field E→\overrightarrow E and magnetic field B→\overrightarrow B . To keep the particle moving in a horizontal direction, it is necessary that (1) B→\overrightarrow B should be perpendicular to the direction of velocity and E→\overrightarrow E should be along the direction of velocity (2) Both B→\overrightarrow B and E→\overrightarrow E should be along the direction of velocity (3) Both B→\overrightarrow B and E→\overrightarrow E are mutually perpendicular and perpendicular to the direction of velocity. (4) B→\overrightarrow B should be along the direction of velocity and E→\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)
Correct Answer
Option C
Solution

Force due to electric field acts along the direction of the electric field but force due to the magnetic field acts along a direction perpendicular to both the velocity of the charged particle and the magnetic field.

Hence both statements (2) and (3) are true.

In statement (2), magnetic force is zero, so, electric force will keep the particle continue to move in horizontal direction.

In statement (3), both electric and magnetic forces will be opposite to each other.

If their magnitudes will be equal then the particle will continue horizontal motion.

Q48
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 ii. The resultant magnetic field due to the two semicircular parts at their common centre is
A μ0i22R{{{\mu _0}i} \over {2\sqrt 2 R}}
B μ0i2R{{{\mu _0}i} \over {2R}}
C μ0i4R{{{\mu _0}i} \over {4R}}
D μ0i2R{{{\mu _0}i} \over {\sqrt 2 R}}
Correct Answer
Option A
Solution

Magnetic fields due to the two parts at their common centre are respectively,

By=μ0i4R{B_y} = {{{\mu _0}i} \over {4R}}

and

Bz=μ0i4R{B_z} = {{{\mu _0}i} \over {4R}}

Resultant field =

By2+Bz2\sqrt {B_y^2 + B_z^2}

=

(μ0i4R)2+(μ0i4R)2\sqrt {{{\left( {{{{\mu _0}i} \over {4R}}} \right)}^2} + {{\left( {{{{\mu _0}i} \over {4R}}} \right)}^2}}
=2μ0i4R=μ0i22R= \sqrt 2 {{{\mu _0}i} \over {4R}} = {{{\mu _0}i} \over {2\sqrt 2 R}}
Q49
A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be
A 900 Ω\Omega
B 1800 Ω\Omega
C 500 Ω\Omega
D 1000 Ω\Omega
Correct Answer
Option A
Solution

Let the resistance to be added be R, then 30 = I g (r + R) ∴\therefore R =

30Ig−r{{30} \over {{I_g}}} - r

=

3030×10−3−100{{30} \over {30 \times {{10}^{ - 3}}}} - 100

= 1000 - 100 = 900

Ω\Omega
Q50
A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is the net force on the remaining three arms of the loop is
A 3F→3\overrightarrow F
B −- F→\overrightarrow F
C −- 3F→3\overrightarrow F
D F→\overrightarrow F
Correct Answer
Option B
Solution

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by

τ=NBiAsin⁡θ\tau = NBiA\sin \theta

. Torque is maximum when θ\theta = 90 o ,i.e., the plane of the coil is parallel to the field

τmax⁡=NBiA{\tau _{\max }} = NBiA

Forces

F→1{\overrightarrow F _1}

and

F→2{\overrightarrow F _2}

acting on the coil are equal in magnitude and opposite in direction. As the forces

F→2{\overrightarrow F _2}

and

F→2{\overrightarrow F _2}

have the same line of action their resultant effect on the coil is zero. The two forces

F→3{\overrightarrow F _3}

and

F→4{\overrightarrow F _4}

are equal in magnitude and opposite in direction.

As the two forces have different lines of action, they constitute a torque.

Thus, if the force on one arc of the loop is

F→\overrightarrow F

, the net force on the remaining three arms of the loop is - F.

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