Revision: Magnetic Fields Due to Electric Current Physics HSC Science (General) 12th Standard Board Exam Maharashtra State Board

When a charged particle moves perpendicular to a uniform magnetic field, it undergoes uniform circular motion.

A moving coil galvanometer is an instrument used to detect and measure small electric currents based on the torque acting on a current-carrying coil placed in a magnetic field.

When a charged particle enters a uniform magnetic field with velocity having both perpendicular and parallel components to the field, it moves in a helical path.

A solenoid is a long, closely wound helical coil of wire that produces a nearly uniform magnetic field inside it when current flows through it.

A toroid is a solenoid bent into a closed circular (ring-shaped) form.

The magnetic dipole moment of a current-carrying coil is defined as the product of the number of turns, the current, and the area of the coil.

μ = N I A

\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}\]

p = qBR

B = \[\frac{\mu_0I}{2\pi d}\]

\[\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi d}\]

K.E. = \[\frac{1}{2}\mathrm{mv}^{2}=\frac{q^{2}B^{2}R_{exit}^{2}}{2m}\]

τ = μB sin θ

τ = NIAB

Deflection Relation:

\[\phi=\frac{NAB}{K}I\]

F = I ∮ dl × B

F = IL × B

\[f_c=\frac{1}{T}=\frac{qB}{2\pi m}\]

F_m​ = q(v × B)

B = \[\frac{\mu_0I}{2R}\]

Magnetic Field at Centre of a Coil (N turns):

B = \[\frac{\mu_0NI}{2R}\]

\[B_z=\frac{\mu_0IR^2}{2(R^2+z^2)^{3/2}}\]

Where:

• I = current
• R = radius of loop
• z = distance of the point from centre along axis
• μ₀ = permeability of free space

Magnetic Field Inside a Long Solenoid:

B = μ_0​ni

Magnetic Field Outside an Ideal Solenoid:

B = 0

B = \[\frac{\mu_0Ni}{2\pi R}\]

F = I ∫ dl × B

τ = μB sin θ

or in vector form,

τ = μ × B

τ = NIAB sin θ

Where:

• N = number of turns
• I = current
• A = area of the loop
• B = magnetic field
• θ = angle between the magnetic field and normal to loop

U = −μ⋅B

Scalar Form: U = −μB cos θ

U_min ​= −μB

U_max​ = +μB

B = \[\frac{\mu_0I\theta}{4\pi r}\]

For θ = 2π,

B = \[\frac{\mu_0I}{2r}\]

The toroid is a solenoid bent into the shape of a hollow doughnut.

According to Ampere's circuital law.

`phivecB.vec(dL) = mu_0I`

Here current 'I' flow through the ring as many times as there are the N no. of turns.

∴ `phivecB.vec(dL) = mu_0NI` ......(1)

Now, B and dL are in the same direction.

∴ `phivecB.vec(dL) = BphidL`

∴ `phivecB.vec(dL) = B.(2pir)` .....(2)

From (1) and (2),

`mu_0NI = B.(2pir)`

∴ B = `(mu_0NI)/(2pir)`

Statement

The magnetic field at a point due to a small current element is directly proportional to the current, the length of the element, and the sine of the angle between the current element and the line joining the element to the point, and inversely proportional to the square of the distance between them.

Mathematical Form

Scalar form:

dB = \[\frac{\mu_0}{4\pi}\frac{Idl\sin\theta}{r^2}\]​

Vector form:

dB = \[\frac{\mu_{0}}{4\pi}\frac{Id\mathbf{l}\times\mathbf{r}}{r^{3}}\]​

where

• μ0​ = permeability of free space
• I = current
• dl = current element
• r = distance from element to point
• θ = angle between dl and r

Explanation

The total magnetic field at a point due to a current-carrying conductor is obtained by integrating (summing) the contributions of all small current elements along the conductor:

B = ∫ dB

Conclusion

Thus, the Biot–Savart Law gives the magnitude and direction of the magnetic field produced by a current-carrying conductor and follows an inverse square law dependence on distance.

Statement

The line integral of the magnetic field around any closed loop is equal to μ₀ times the net current enclosed by the loop.

∮ B ⋅ dl = μ₀I_enc​

where

• B = magnetic field
• dl = small element of the closed loop
• I_enc = net current enclosed
• μ_0​ = permeability of free space

Explanation / Proof 

Consider a long straight wire carrying current I.

Due to cylindrical symmetry:

• Magnetic field B is tangential to a circular path around the wire.
• Magnitude of B is the same at all points on a circle of radius r.

Choose a circular Amperian loop of radius r.

Since B and dl are parallel:

∮ B ⋅ dl = ∮ B dl

=B∮dl

=B(2πr)

By Ampere’s Law:

B(2πr) = μ₀I

B = \[\frac{\mu_0I}{2\pi r}\]​

This matches the magnetic field obtained earlier.

Conclusion

Hence,

∮ B ⋅ dl = μ₀I_enc​

is verified and is known as Ampere’s Circuital Law, a fundamental law of magnetostatics.