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NEET Exam  >  NEET Tests  >  Test: Torque on Current Loop Magnetic Dipole (December 30) - NEET MCQ

Test: Torque on Current Loop Magnetic Dipole (December 30) - NEET MCQ


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10 Questions MCQ Test - Test: Torque on Current Loop Magnetic Dipole (December 30)

Test: Torque on Current Loop Magnetic Dipole (December 30) for NEET 2024 is part of NEET preparation. The Test: Torque on Current Loop Magnetic Dipole (December 30) questions and answers have been prepared according to the NEET exam syllabus.The Test: Torque on Current Loop Magnetic Dipole (December 30) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Torque on Current Loop Magnetic Dipole (December 30) below.
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Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 1
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A uniform conducting wire of length length 10a and resistance R is wound up into four turn as a current carrying coil in the shape of equilateral triangle of side a. If current I is flowing through the coil then the magnetic moment of the coil is
image

Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 1

Magnetic moment M = NIA
= 4 × I × 
= √3a2I 

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 2
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A moving coil galvanometer can be converted into an ammeter by:

Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 2

To utilise a galvanometer (G) as an ammeter, there is the difficulty in measurement of current due to the sensitivity of galvanometer and also the connection of galvanometer with a very large resistance in series that may change the value of current in the circuit.
To overcome these difficulties one attaches a small resistance rs called shunt resistance in parallel with the galvanometer coil as shown in the figure.
solution

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Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 3
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A galvanometer of resistance 70Ω, is converted to an ammeter by a shunt resistance rs = 0.03Ω. The value of its resistance will become

Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 3


Here, RG = 70Ω, rs = 0.03Ω
∴ R = 
= 0.02998 
= 0.03Ω

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 4
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A  circular coil of radius 10cm having 100 turns carries a current of 3.2A. The magnetic field at the center of the coil is
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 4

As 
Here N = 100,
I = 3.2A,
R = 10cm = 10 × 10−2m
∴ B = 
= 2.01 × 10−3T

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 5
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A galvanometer having a resistance of 50Ω, gives a full scale deflection for a current of 0.05A. The length (in metres) of a resistance wire of area of cross section 3 × 10−2cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is
(Specific resistance of the wire ρ = 5 × 10−7Ωm)
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 5


= 
= 
= 
= 
∵ S = 
∴ l  = 
= 3.0m

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 6
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A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of 30o with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 6

N = 70, r = 5cm = 5 × 10−2m,I = 8A 
B = 1.5T, θ = 30∘
The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil,
∴ τ = NIABsinθ = NIπr2Bsin30∘ 
= 70 × 8 × 3.14 × (5 × 10−2)2 × 1.5 × 1/2
= 3.297 N m
≈ 3.3 N m

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 7
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A galvanometer of resistance 10Ω gives full-scale deflection when 1mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5V is
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 7

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 8
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In a moving coil galvanometer the deflection (ϕ) on the scale by a pointer attached to the spring is
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 8

Since magnetic torque on the coil,
τ = NIAB
This torque is balanced by counter torque
∴ kϕ = 
where k is torsional constant. It is a scalar quantity having dimension of torque or energy i.e. [ML2T−2]

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 9
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A short bar magnet has a magnetic moment of 0.65JT−1, then the magnitude and direction of the magnetic field produced by the magnet at a distance 8cm from the centre of magnet on the axis is
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 9

Here, M = 0.65JT−1,
d = 8cm
= 0.08m
The field produced by magnet at axial point is given by
B = 
= 2.5 × 10−4T along SN

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 10
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The torque on a current-carrying rectangular loop is τ when it is placed in a uniform magnetic field at an angle θ. If the length and the width of the loop is doubled then the torque on the loop at an angle θ will be:"
Detailed Solution for Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 10

Given:

  • The torque (τ) on a rectangular loop is proportional to the area of the loop (A), the current (I), the magnetic field strength (B), and sinθ:
    τ=nIABsinθ
    where:
    • n = number of turns (assume n=1 for simplicity)
    • I = current
    • A = length × width = area of the loop
    • B = magnetic field strength
    • θ = angle between the normal to the plane of the loop and the magnetic field.

Initial Area:

Let the original length of the loop be l and the width be w. Thus:

When Length and Width are Doubled:

The new length becomes 2l, and the new width becomes 2w. Therefore, the new area of the loop is:

Anew = (2l) ⋅ (2w) = 4 ⋅ (l⋅w) = 4Ainitial

Torque on the New Loop:

Since the torque is directly proportional to the area of the loop, doubling both the length and width results in a torque that is 4 times the original torque:

τnew ​= 4⋅τinitial​

Final Answer:

The torque on the loop at an angle θ will be:
τnew​ = 4τ

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