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The magnetic moment of a current I carrying circular coil of radius r and number of turns N varies as
Magnetic Moment, M = NIA = NIπr^{2}
i.e. M ∝ r^{2}
A 200 turn closely wound circular coil of radius 15 cm carries a current of 4 A. The magnetic moment of this coil is:
The magnetic moment is given by
= 200 × 4 × 3.14 × (15 × 10^{−2})^{2}
= 200 × 4 × 3.14 × 15 × 15 × 10^{4}
= 56.5 A m^{2}
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
Magnetic moment M = NIA
= 4 × I ×
= √3a^{2}I
The magnetic moment associated with a circular coil of 35 turns and radius 25cm, if it carries a current of 11A is
Given N = 35,
r = 25cm
= 25 × 10^{−2}m
I = 11A
Then magnetic moment associated with this circular coil
M = NIA = NIπr^{2} = 35 × 11 × 3.14 × (25 × 10^{−2})^{2}
= 75.56 A m^{2}
A uniform conducting wire of length 18a and resistance R is wound up as current carrying coil in the shape of a regular hexagon of sides a. If the coil is connected to a voltage source V_{0}, then the magnetic moment of coil is
In the regular hexagon, if each arm length is a,
then the number of turns in the given shape, n = 18a / 6a = 3
Now the area of given shape is
Now magnetic moment M = nIA
=
=
The magnetic field at the centre of a circular loop of area A is B. The magnetic moment of the loop is
Let r be the radius of the circular loop
Magnetic field at the centre of the loop is
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
Here, M = 0.65JT^{−1},
d = 8cm
= 0.08m
The field produced by magnet at axial point is given by
B =
= 2.5 × 10^{−4}T along SN
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
As
Here N = 100,
I = 3.2A,
R = 10cm = 10 × 10^{−2}m
∴ B =
= 2.01 × 10^{−3}T
A circular coil of radius 10cm having 100 turns carries a current of 3.2A. For the coil given the magentic moment is
The magentic moment is given by
m = NIA = NIπr^{2} = 100 x 3.2 x 3.14 x (10 x 10^{2})^{2}
= 100 x 3.2 x 3.14 x 10^{2}
= 10 A m^{2}
A circular coil of radius 10cm having 100 turns carries a current of 3.2A. the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of 60∘ under the influence of magnetic field. The magnitude of torque on the coil in the final position is
Torque,
Here, m = 10Am^{2}, B=5T
Now initially θ = 0^{∘}
Thus, initial torque, τ_{i} =0
In final position θ = 60^{∘}
∴ τ_{f} = mBsin60^{∘ }=
= 25√3 N m
A current carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend upon
The torque depends upon n, I, A, B and θ. But it does not depend upon the shape of the loop (rectangular, circular, triangular, etc.)
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 30^{o} with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is
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 = NIπr^{2}Bsin30^{∘}
= 70 × 8 × 3.14 × (5 × 10^{−2})^{2} × 1.5 × 1/2
= 3.297 N m
≈ 3.3 N m
A circular coil of 25 turns and radius 12 cm is placed in a uniform magnetic field of 0.5 T normal to the plane ol the coil. If the current in the coil is 6 A then total torque acting on the coil is
The torque acting on the coil
Here the circular coil is placed normal to the direction of magnetic field then the angle between the direction of magnetic moment is zero, then
τ = mBsinθ = mBsin0 = 0
∴ τ = 0
The final torque on a coil having magnetic moment 25 A m^{2} in a 5T uniform external magnetic field, if the coil rotates through an angle of 60^{∘} under the influence of the magnetic field is
Here, m = 25Am^{2}; θ = 60^{∘}; B = 5T
∴ τ = 25 × 5 × sin60^{∘} or
τ =
= 108.25 N m
The magnitude of torque experienced by a square coil of side 12cm which consists of 25 turns and carries a current 10A suspended vertically and the normal to the plane of coil makes an angle of 30^{∘} with the direction of a uniform horizontal magnetic field of magnitude 0.9T is
τ = NIABsinθ
Here, N = 25,
I = 10A,
B = 0.9T,
θ = 30^{∘}
A = a^{2}^{ }= 12 × 10^{−2} × 12 × 10^{−2 }
144 × 10^{−4}m^{2}
∴ τ = 25 × 10 × 144 × 10^{−4} × 0.9 × sin30^{∘} = 1.6 Nm
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