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 Daily Test for NEET Preparation - Test: Torque on Current Loop Magnetic Dipole (December 30)

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

### 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

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

Magnetic moment M = NIA
= 4 × I ×
= √3a2

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

### 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 (December 30) - 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 (December 30) - Question 3

### A moving coil galvanometer can be converted into an ammeter by:

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

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 4

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 4

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

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 5

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 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 (December 30) - Question 6

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 6

∵ S =
∴ l  =
= 3.0m

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 7

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 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 (December 30) - Question 8

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 8

Test: Torque on Current Loop Magnetic Dipole (December 30) - Question 9

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 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 (December 30) - Question 10

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 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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