Learn Unit 3-Magnetic Effect of Current & Magnetism with Free Lessons & Tips

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:

• dBdB is the infinitesimal magnetic field vector,
• μ0μ0 is the permeability of free space,
• II is the current flowing through the element,
• dldl is the infinitesimal current element vector,
• r^r^ is the unit vector pointing from the current element to the point where the magnetic field is being calculated,
• rr is the distance from the current element to the point where the magnetic field is being calculated.

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.

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:

• FF is the magnetic force,
• qq is the charge of the particle,
• vv is the velocity of the particle,
• BB is the magnetic field strength.

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.

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:

• BPBP is the magnetic field produced by loop P at point O,
• BQBQ is the magnetic field produced by loop Q at point O.

The formula for the magnetic field at the center of a circular loop is given by:

B=μ0I2rB=2rμ0I

Where:

• μ0μ0 is the permeability of free space (4π×10−7 T⋅m/A4π×10−7T⋅m/A),
• II is the current flowing through the loop,
• rr is the radius of the loop.

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.

(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:

• BB is the magnetic field at the center of the coil,
• μ0μ0 is the permeability of free space (4π×10−7 T⋅m/A4π×10−7T⋅m/A),
• NN is the number of turns in the coil,
• II is the current flowing through the coil,
• rr is the radius of the coil.

(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:

• μμ is the magnetic moment of the coil,
• NN is the number of turns in the coil,
• II is the current flowing through the coil,
• rr is the radius of the coil.

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:

• BB as the magnetic field inside the solenoid,
• μμ as the permeability of the medium,
• II as the current passing through the solenoid,
• NN as the number of turns per unit length of the solenoid.

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

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:

• BB as the magnetic field inside the toroid,
• μμ as the permeability of the medium,
• II as the current passing through the toroid,
• NN as the total number of turns of wire in the toroid,
• rr as the radius of the toroid,
• RR as the radius of the cross-section of the toroid.

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.