JEE Mains Physics · Formula sheet

Moving Charges and Magnetism formulas

16 formulas, 2 reference tables and 54 common traps for JEE Mains Physics Moving Charges and Magnetism, grouped by subtopic.

Full notes with worked examples

Field of Straight Wires and Arcs

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Field of a long straight wire and of two parallel wires

Long straight wire

B = \frac{\mu_0 I}{2\pi d}

Field of a finite straight wire and of a polygon loop

Finite straight wire

B = \frac{\mu_0 I}{4\pi d}\left(\sin\alpha_1 + \sin\alpha_2\right) \qquad dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^{2}}

Field at the centre of arcs and bent wires

Piece of wire	Field at the point	For I = 10 A at a distance or radius of 5 cm
Infinite straight wire, point at distance d	$\dfrac{\mu_0 I}{2\pi d}$	$40\ \mu\text{T}$
Semi-infinite wire, point on the perpendicular through its end	$\dfrac{\mu_0 I}{4\pi d}$	$20\ \mu\text{T}$ Half of the infinite wire, not the same.
Straight wire whose line passes through the point	Zero	Zero
Full circular loop, at its centre	$\dfrac{\mu_0 I}{2R}$	$40\pi\ \mu\text{T} \approx 126\ \mu\text{T}$
Three-quarter circle, at its centre	$\dfrac{3\mu_0 I}{8R}$	$30\pi\ \mu\text{T} \approx 94.2\ \mu\text{T}$
Semicircle, at its centre	$\dfrac{\mu_0 I}{4R}$	$20\pi\ \mu\text{T} \approx 62.8\ \mu\text{T}$
Quarter circle, at its centre	$\dfrac{\mu_0 I}{8R}$	$10\pi\ \mu\text{T} \approx 31.4\ \mu\text{T}$
Arc of angle θ in radians, at its centre	$\dfrac{\mu_0 I\theta}{4\pi R}$	$20\theta\ \mu\text{T}$

Every piece's field at the centre is perpendicular to the plane of the wire, so in a flat shape the pieces add or subtract along one line.

Common traps

Opposite currents add between the wires

Between two antiparallel currents both fields point the same way, so with equal currents the midpoint field is twice one wire's field. Subtracting them gives zero, which is the answer only for like currents.

Outside the pair the rule reverses

At a point beyond both wires, antiparallel currents give opposite fields and like currents give fields in the same direction. Draw each field's direction at the point before adding.

Fields at an angle add as vectors

When the point is not on the line of the wires, the two fields are not parallel. If the lines from the point to the wires are perpendicular, B = √(B₁² + B₂²), not B₁ + B₂.

The angles are measured from the perpendicular

In B = (μ₀I/4πd)(sin α₁ + sin α₂), each α is the angle between the perpendicular from the point and the line to an end of the wire. Measuring it from the wire swaps sine for cosine.

A semi-infinite wire gives half, not the full value

At a point level with the end of a semi-infinite wire, the field is μ₀I/4πd. Using μ₀I/2πd, the infinite-wire result, doubles that piece's contribution.

The centre of a polygon is not at a distance a/2

For a triangle the centre is a/2√3 from each side; only for a square is it a/2. The general distance is a/(2 tan(π/n)).

A straight piece aimed at the centre gives nothing

Radial straight pieces, and straight leads whose line passes through the centre, contribute zero. Leaving them out is correct; giving them μ₀I/4πd is not.

The angle of an arc must be in radians

B = μ₀Iθ/4πR uses θ in radians. A 90° arc is θ = π/2, giving μ₀I/8R; putting θ = 90 gives a field larger by a factor of about 57.

Check the sense of every piece before adding

Two arcs, or an arc and a straight lead, can give fields in opposite directions at the centre. Decide into or out of the page for each piece first; adding the sizes alone is the most common wrong option.

Circular Loops and Coils

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Field at the centre of a circular coil

Centre of a coil and energy density

B = \frac{\mu_0 N I}{2R} \qquad u = \frac{B^{2}}{2\mu_0}

Field on the axis of a circular loop

Axis of a loop

B = \frac{\mu_0 N I R^{2}}{2\left(R^{2} + x^{2}\right)^{3/2}} \qquad \frac{B}{B_c} = \frac{R^{3}}{\left(R^{2} + x^{2}\right)^{3/2}}

Common traps

Rewinding changes the radius too

The same wire made into more turns makes each turn smaller. B depends on N/R and R goes as 1/N, so B goes as N². Scaling only by N misses half the change.

Perpendicular coils add as vectors

Two coils in perpendicular planes give fields along two perpendicular axes. The net field is √(B₁² + B₂²), not B₁ + B₂.

Do not forget the number of turns

μ₀I/2R is one turn. A coil of N closely wound turns gives N times that; leaving out N is a common slip when the turns are given in a separate sentence.

The power is 3/2, not 1/2

The axis field has (R² + x²)^(3/2) in the denominator and R² on top. Using the square root, or leaving out R², gives an expression with the wrong units.

An approximation needs x much less than R

A binomial approximation of the axis field is valid only very close to the centre. When x is comparable with R, work out (R/√(R² + x²))³ exactly.

The direction of each coaxial loop's field

Seen from the point between two loops, a clockwise current gives a field pointing towards that loop. Two loops whose currents both look clockwise from the point give opposite fields there, which subtract.

Ampère's Law: Thick Wires and Solenoids

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Ampère's law for a solid wire, a hollow tube and a coaxial cable

Ampère's law and a solid wire

\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enc}} \qquad B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^{2}},\quad B_{\text{out}} = \frac{\mu_0 I}{2\pi r}

Field inside a solenoid and a toroid

Solenoid, intensity, core and toroid

B = \mu_0 n I \qquad H = nI \qquad B = \mu_0\mu_r n I \qquad B_{\text{toroid}} = \frac{\mu_0 N I}{2\pi r}

Common traps

Inside a solid wire B grows with r

Inside a wire with uniform current, B = μ₀Ir/2πa², which rises from zero at the axis. Using μ₀I/2πr inside gives a field that blows up at the axis, the opposite shape.

The field outside a coaxial cable is zero

With equal and opposite currents in the inner wire and the shell, a circle outside both encloses no net current, so B = 0 there. Between the conductors only the inner current counts.

Only enclosed current counts

Current flowing just outside an Amperian loop changes B at points on the loop but not the line integral of B round it. The integral is μ₀ times the current that passes through the loop.

Turns per centimetre must become turns per metre

In B = μ₀nI, n is per metre. A winding of 20 turns per cm is n = 2000 per metre; leaving it as 20 makes B a hundred times too small.

H has no μ₀ in it

Magnetic intensity inside a solenoid is H = nI, in A/m. B = μ₀nI is in tesla. A question that asks for magnetic intensity wants H.

Flux and flux linkage differ by N

The flux through one cross-section of a solenoid is BA. The flux linked with the whole winding is NBA. The two answers differ by the number of turns, and both appear among the options.

Lorentz Force and Crossed Fields

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Magnetic force on a moving charge, F = q v × B

Magnetic force

\vec F = q\left(\vec v \times \vec B\right), \qquad F = qvB\sin\theta

Electric and magnetic forces together: the velocity selector

Lorentz force and the selected speed

\vec F = q\left(\vec E + \vec v \times \vec B\right) \qquad v = \frac{E}{B}

Common traps

v × B is not B × v

The cross product changes sign when the order is swapped. F = q v × B; writing B × v gives the force in exactly the opposite direction.

An electron's force is reversed

For an electron q = −e, so the force is along −(v × B). Finding the direction of v × B and stopping there gives the wrong answer for every electron question.

The magnetic force never changes speed

Because F is always perpendicular to v, it does no work. Speed and kinetic energy stay the same; only the direction of motion changes.

The selector picks a speed, not a charge or a mass

v = E/B has no q and no m in it. A proton, an electron and an alpha particle at the same speed all pass straight through the same crossed fields.

Find v from the kinetic energy first

When a question gives an energy in eV, convert it to joules and use v = √(2K/m) before putting v into E = vB. Using the energy directly as a speed gives nonsense.

Constant velocity is impossible with only an electric field

An electric field always pushes a charge, whatever its motion. A region with E ≠ 0 and B = 0 cannot let a charge move at constant velocity; a region with B alone can, if the charge moves along B.

Circular and Helical Paths of Charges

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Radius of a charge's circular path and how it compares between particles

Radius of the circle

r = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}

Finding a radius, a mass or a field from r = mv/qB

Mass from a measured radius

m = \frac{qBr}{v} = \frac{qB^{2}r^{2}}{2V}

Period of revolution, the helix and the cyclotron

Period, pitch and cyclotron energy

T = \frac{2\pi m}{qB} \qquad p = v\cos\theta\,T \qquad K_{\max} = \frac{q^{2}B^{2}R^{2}}{2m}

Common traps

Equal energy and equal voltage are different conditions

At equal kinetic energy r goes as √m/q; after the same accelerating voltage r goes as √(m/q). For an alpha particle against a proton the two give different ratios, so read which one the question fixes.

Curvature is not radius

Curvature is 1/r. A particle with a larger radius has a smaller curvature and is deflected less. Reading 'smaller curvature' as 'smaller radius' flips the comparison.

Radius goes as the square root of kinetic energy

r = √(2mK)/qB, so doubling K multiplies r by √2, not 2. The radius is proportional to momentum, not to energy.

Energy in eV must be turned into joules

Put kinetic energy into √(2mK) in joules: 1 eV = 1.6 × 10⁻¹⁹ J. A keV is a thousand of those.

Radius, not diameter

When a charge goes round half a circle and exits, the gap between entry and exit is 2r. Putting that gap into r = mv/qB doubles the mass or halves the field.

Singly ionised means charge e

A singly ionised atom has q = e whatever its mass number; the mass number gives only the mass, in u.

Only the part of v along B makes the pitch

Pitch is v cos θ times the period. Using the full speed, or the part across B, gives a wrong distance along the field.

Two gains of energy per revolution in a cyclotron

The particle crosses the gap between the dees twice in each turn, gaining qV each time. Revolutions = K/(2qV); dividing by qV alone doubles the count.

The period does not depend on the speed

T = 2πm/qB has no v and no r in it. A faster charge goes round a larger circle in the same time, which is why the cyclotron frequency can stay fixed.

Forces and Torques on Current-Carrying Wires

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Force on a current-carrying wire in a magnetic field

Force on a wire

\vec F = I\,\vec L \times \vec B, \qquad F = ILB\sin\theta

Force between two parallel currents

Force per metre between parallel wires

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Magnetic moment of a coil and the torque on it

Moment and torque

\vec m = NI\vec A, \qquad \vec \tau = \vec m \times \vec B, \qquad \tau = NIAB\sin\theta

Common traps

Only the part in the field counts

When a wire runs partly through a field region, L in ILB is the length inside the region. Using the whole wire overstates the force.

A bent wire's force uses the chord

In a uniform field, a semicircle of radius R feels the same force as a straight wire of length 2R joining its ends, not πR.

θ is the angle between the wire and the field

F = ILB sin θ is largest when the wire is perpendicular to B and zero when the wire lies along B. A wire at 30° to the field feels half the largest force.

Like currents attract

This is the opposite of charges: two wires with current in the same direction pull together. Opposite currents push apart.

The two forces are equal even if the currents differ

A 2 A wire beside a 10 A wire feels the same size of force as the 10 A wire does, over the same length. The force depends on the product I₁I₂, which is the same for both.

The force goes as the product, not the sum

F/L = μ₀I₁I₂/2πd. Doubling both currents multiplies the force by four, and halving the distance doubles it again.

θ is measured from the axis, not the plane

In τ = NIAB sin θ, θ is between the coil's normal and B. If a question gives the angle between the plane and the field, use 90° minus it.

Opposite currents give opposite moments

Two concentric loops with currents in opposite senses have moments pointing opposite ways. The net moment is the difference, not the sum.

Do not forget N

A coil of N turns has N times the moment of one turn. The torque, NIAB sin θ, carries the same factor N.

Galvanometer, Ammeter and Voltmeter

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Moving coil galvanometer: deflection and sensitivity

Balance and sensitivity

NIAB = C\theta \qquad \frac{\theta}{I} = \frac{NAB}{C} \qquad \frac{\theta}{V} = \frac{NAB}{CR}

Converting a galvanometer into an ammeter with a shunt

Shunt

S = \frac{I_g G}{I - I_g} \qquad R_A = \frac{SG}{S + G}

Converting a galvanometer into a voltmeter, compared with an ammeter

Property	Ammeter	Voltmeter
What it measures	Current, up to I	Potential difference, up to V
Resistance added	Shunt $S = \dfrac{I_g G}{I - I_g}$	Series $R = \dfrac{V}{I_g} - G$
How it is joined to the galvanometer	In parallel	In series
How the meter goes into the circuit	In series with the part	In parallel, across the part Mirror images: swap both connections when you swap meters.
Resistance of the finished meter	$\dfrac{SG}{S + G}$ , less than S	$G + R$ , large
Ideal resistance	Zero	Infinite
Range made n times larger	Shunt $\dfrac{G}{n - 1}$	Add $(n - 1)$ times the meter's resistance in series
For G = 100 Ω and Ig = 1 mA	1 A range: S ≈ 0.1 Ω	10 V range: R = 9900 Ω

A small resistance in parallel for current; a large resistance in series for voltage.

Common traps

More turns also mean more resistance

Adding turns of the same wire raises NAB and R together. Current sensitivity grows, but voltage sensitivity, NAB/CR, does not change.

Figure of merit is the inverse of sensitivity

Current sensitivity is divisions per ampere; figure of merit is amperes per division. A more sensitive galvanometer has a smaller figure of merit.

Current and voltage sensitivity differ by R

θ/V = (θ/I)/R. Comparing two galvanometers' voltage sensitivity needs each one's resistance as well as N, A, B and C.

The shunt carries I − Ig, not I

S = IgG/(I − Ig). Using I in the denominator gives a slightly smaller shunt, and with a small range like 5 mA against 1 mA the error is large.

A shunt goes in parallel

The shunt is joined across the galvanometer. A resistor in series raises the range for voltage, not current.

The ammeter's resistance is less than the shunt

The coil and shunt are in parallel, so the meter's resistance SG/(S + G) is below both. It is not S + G.

Subtract the galvanometer's own resistance

The series resistor is V/Ig − G. V/Ig alone is the whole voltmeter's resistance, coil included.

Raising a voltmeter's range uses the meter's resistance

To multiply the range by n, add (n − 1) times the voltmeter's total resistance, not (n − 1) times the bare galvanometer's.

A voltmeter in series reads almost the whole supply

Its large resistance takes nearly all the voltage and lets almost no current through. A voltmeter must go across the part, in parallel.

Moving Charges and Magnetism formulas

Field of Straight Wires and Arcs

Field of a long straight wire and of two parallel wires

Field of a finite straight wire and of a polygon loop

Field at the centre of arcs and bent wires

Common traps

Circular Loops and Coils

Field at the centre of a circular coil

Field on the axis of a circular loop

Common traps

Ampère's Law: Thick Wires and Solenoids

Ampère's law for a solid wire, a hollow tube and a coaxial cable

Field inside a solenoid and a toroid

Common traps

Lorentz Force and Crossed Fields

Magnetic force on a moving charge, F = q v × B

Electric and magnetic forces together: the velocity selector

Common traps

Circular and Helical Paths of Charges

Radius of a charge's circular path and how it compares between particles

Finding a radius, a mass or a field from r = mv/qB

Period of revolution, the helix and the cyclotron

Common traps

Forces and Torques on Current-Carrying Wires

Force on a current-carrying wire in a magnetic field

Force between two parallel currents

Magnetic moment of a coil and the torque on it

Common traps

Galvanometer, Ammeter and Voltmeter

Moving coil galvanometer: deflection and sensitivity

Converting a galvanometer into an ammeter with a shunt

Converting a galvanometer into a voltmeter, compared with an ammeter

Common traps

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