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JEE Exam  >  JEE Tests  >  Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - JEE MCQ

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - JEE MCQ


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10 Questions MCQ Test - Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec)

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) for JEE 2024 is part of JEE preparation. The Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) questions and answers have been prepared according to the JEE exam syllabus.The Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) below.
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Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - 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 The Moving Coil Galvanometer(5 Dec) - Question 1

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

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 2
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The magnetic moment associated with a circular coil of 35 turns and radius 25cm, if it carries a current of 11A is

Detailed Solution for Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 2

Given N = 35,
r = 25cm
= 25 × 10−2m
I = 11A
Then magnetic moment associated with this circular coil
M = NIA = NIπr2 = 35 × 11 × 3.14 × (25 × 10−2)2
= 75.56 A m2

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Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 3
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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 The Moving Coil Galvanometer(5 Dec) - Question 3

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

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 4
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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 The Moving Coil Galvanometer(5 Dec) - Question 4


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

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 5
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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 The Moving Coil Galvanometer(5 Dec) - Question 5

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

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 6
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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 The Moving Coil Galvanometer(5 Dec) - Question 6


= 
= 
= 
= 
∵ S = 
∴ l  = 
= 3.0m

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 7
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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 The Moving Coil Galvanometer(5 Dec) - Question 7

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 The Moving Coil Galvanometer(5 Dec) - Question 8
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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 The Moving Coil Galvanometer(5 Dec) - Question 8

Test: Torque on Current Loop, Magnetic Dipole The Moving Coil Galvanometer(5 Dec) - Question 9
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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 The Moving Coil Galvanometer(5 Dec) - Question 9

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 The Moving Coil Galvanometer(5 Dec) - Question 10
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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 The Moving Coil Galvanometer(5 Dec) - Question 10

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

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