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Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
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In this diagram, the loop carrying current is represented by the straight lines forming the loop. The magnetic field lines form concentric circles around the loop, indicating the direction of the magnetic field. According to the right-hand rule, the direction of the magnetic field lines can be determined by curling the fingers of the right hand in the direction of the current flow around the loop. The thumb then points in the direction of the magnetic field lines inside the loop.
read lessAnswered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
To generalize Ampere's circuital law to include the term due to displacement current, we use Maxwell's modification of Ampere's law, which states:
∮B⋅dl=μ0(∬J⋅dA+ε0dtd∬E⋅dA)
Where:
This modification includes the term ε0ddt∬E⋅dAε0dtd∬E⋅dA, which accounts for the displacement current.
read lessAnswered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
Biot-Savart's law states that the magnetic field BB at a point due to an infinitesimal current element dldl carrying current II at a distance rr from the element is given by:
dB=μ04πI dl×r^r2dB=4πμ0r2Idl×r^
Where:
In vector form, Biot-Savart's law can be expressed as:
B=μ04π∫I dl×r^r2B=4πμ0∫r2Idl×r^
This equation represents the vector sum of the magnetic fields contributed by all infinitesimal current elements along the path of integration.
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Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
The expression for the magnitude of the magnetic field at the center of a circular loop of radius rr carrying a steady current II can be derived using Ampere's law or Biot-Savart's law. For a circular loop, the magnetic field at its center is given by:
B=μ0I2rB=2rμ0I
Where:
The direction of the magnetic field at the center of the loop is perpendicular to the plane of the loop and is along the axis passing through the center of the loop.
Now, let's draw the magnetic field lines due to the current loop:
____B____
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In this diagram, the loop is represented by the circle with current flowing clockwise. The magnetic field lines form concentric circles around the loop, with their direction indicated by arrows. At the center of the loop, the magnetic field lines are directed perpendicular to the plane of the loop.
Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
One tesla (1 T) is defined as the magnetic field that exerts a force of one newton (1 N) on a particle of charge qq moving with velocity vv perpendicular to the magnetic field BB. Mathematically, this can be expressed using the formula for the magnetic force (FF) acting on a charged particle:
F=qvBF=qvB
Where:
Therefore, when q=1q=1 coulomb and v=1v=1 meter per second, and the magnetic force exerted is 11 newton, the magnetic field strength BB is 11 tesla.
Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
To find the magnitude of the net magnetic field at point O due to the two identical circular loops P and Q, we can use the superposition principle, which states that the total magnetic field at a point is the vector sum of the magnetic fields produced by individual current elements.
Since the loops are identical and carry equal currents, the magnetic fields produced by each loop at point O will have the same magnitude but opposite directions due to the currents flowing in opposite directions.
Therefore, the net magnetic field at point O is the difference between the magnetic fields produced by the two loops:
Bnet=BP−BQBnet=BP−BQ
Where:
The formula for the magnetic field at the center of a circular loop is given by:
B=μ0I2rB=2rμ0I
Where:
Since both loops have the same radius and carry equal currents, we can simplify the expression for the net magnetic field:
Bnet=μ0I2r−μ0I2rBnet=2rμ0I−2rμ0I Bnet=0Bnet=0
Therefore, the magnitude of the net magnetic field at point O is zero.
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Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
(i) The expression for the magnetic field at the center of a circular coil carrying current II with NN turns and radius rr is given by:
B=μ0NI2rB=2rμ0NI
Where:
(ii) The magnetic moment (μμ) of the coil is given by the product of the current II and the area enclosed by the coil. For a circular coil, the area (AA) enclosed by the coil is given by πr2πr2. Therefore, the expression for the magnetic moment (μμ) is:
μ=NI⋅Aμ=NI⋅A
μ=NI⋅πr2μ=NI⋅πr2
μ=Nπr2Iμ=Nπr2I
Where:
Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
When a charged particle moves with velocity vv perpendicular to a uniform magnetic field BB, it experiences a magnetic force (FBFB) perpendicular to both vv and BB. According to the Lorentz force law, the magnetic force (FBFB) acting on the particle is given by:
FB=qv×BFB=qv×B
Since the force is always perpendicular to the velocity, it changes the direction of the velocity but not its magnitude. This results in the particle moving along a circular path.
To find the radius of the circular path, we equate the magnetic force (FBFB) to the centripetal force (FcFc) required to keep the particle in circular motion:
qvB=mv2rqvB=rmv2
Where:
From this equation, we can solve for the radius rr of the circular path:
r=mvqBr=qBmv
Therefore, the expression for the radius rr of the circular path is given by:
r=mvqBr=qBmv
Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
To derive the expression for the magnetic field inside a long solenoid, we can use Ampere's law. Ampere's law states that the line integral of the magnetic field (BB) around a closed path is equal to the permeability of the medium (μμ) times the total current passing through the surface enclosed by the path.
For a long solenoid, the magnetic field inside is essentially uniform and parallel to the axis of the solenoid. Therefore, we can choose a rectangular path parallel to the axis of the solenoid to apply Ampere's law.
Let's denote:
Considering a rectangular path with sides parallel to the solenoid axis, the line integral of BB along the sides parallel to the axis is zero since the magnetic field is parallel to the sides.
The line integral of BB along the sides perpendicular to the axis (top and bottom) is:
∮B⋅dl=B⋅2l∮B⋅dl=B⋅2l
where ll is the length of the solenoid.
By Ampere's law, this is equal to μμ times the total current passing through the surface enclosed by the path. Since the current is uniform throughout the solenoid, the total current passing through the surface is I⋅AI⋅A, where AA is the cross-sectional area of the solenoid.
Therefore, we have:
B⋅2l=μIAB⋅2l=μIA
From this equation, we can solve for BB, the magnetic field inside the solenoid:
B=μI2lAB=2lμIA
The cross-sectional area AA of the solenoid is equal to the area of each turn multiplied by the number of turns per unit length (NN):
A=πr2⋅NA=πr2⋅N
Substituting this expression for AA into the equation for BB, we get:
B=μI2lπr2⋅NB=2lμIπr2⋅N
Therefore, the expression for the magnetic field inside the solenoid is:
B=μI2⋅NB=2μI⋅N
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Answered on 06 Apr Learn Unit 3-Magnetic Effect of Current & Magnetism
Sadika
To obtain the expression for the magnetic field inside a toroid carrying current, let's consider a toroidal solenoid, which is a hollow circular ring wound uniformly with NN turns of wire carrying a current II.
Let's denote:
Now, let's consider a circular path inside the toroid, which is concentric with the toroid itself.
The line integral of BB along this circular path is equal to the permeability of the medium times the total current passing through the surface enclosed by the path.
∮B⋅dl=μIenc∮B⋅dl=μIenc
Where IencIenc is the total current passing through the surface enclosed by the circular path.
Since the current is uniformly distributed along the wire of the toroid, the total current passing through the circular path is equal to the current II multiplied by the total number of turns NN.
Ienc=NIIenc=NI
Therefore, we have:
∮B⋅dl=μNI∮B⋅dl=μNI
By symmetry, the magnetic field BB at every point along the circular path is tangential to the path and has the same magnitude.
The line integral ∮B⋅dl∮B⋅dl along the circular path is equal to BB multiplied by the circumference of the circular path.
B⋅2πr=μNIB⋅2πr=μNI
From this equation, we can solve for BB, the magnetic field inside the toroid:
B=μNI2πrB=2πrμNI
Therefore, the expression for the magnetic field inside the toroid is:
B=μNI2πrB=2πrμNI
This expression shows that the magnetic field inside the toroid is inversely proportional to the radius rr of the circular path and directly proportional to the total current II passing through the toroid and the number of turns NN.
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