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Magnetic Fields Due to Electric Current formulas

10 formulas and 12 common traps for MHT-CET Physics Magnetic Fields Due to Electric Current, grouped by subtopic.

Full notes with worked examples

Force on a Moving Charge, and Circular Motion in a Field

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The Magnetic Force on a Moving Charge

Lorentz force

F⃗=qE⃗+q(v⃗×B⃗)\vec F = q\vec E + q(\vec v \times \vec B)

Circular Motion in a Magnetic Field

Radius

r=mvqB=2mKqBr = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}

Common traps

Thinking a magnetic field can change a charge's speed

The magnetic force is perpendicular to the velocity at every instant, so it does no work: kinetic energy and speed stay fixed, only the direction turns.

Scaling the radius with the energy

r ∝ √K, not K. Doubling the kinetic energy makes the circle √2 times wider, not twice.

Force on a Current-Carrying Wire, and Between Parallel Wires

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Force on a Straight Conductor

Force on a wire

F⃗=I L⃗×B⃗\vec F = I\,\vec L \times \vec B

Parallel Wires: Attraction, Repulsion and Balance

Force between parallel wires

Fl=μ0I1I22πd\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}

Common traps

Expecting the near and far sides to cancel

They carry opposite currents but sit at different distances from the wire, where the field differs. Only the perpendicular sides cancel; the net force is the difference of the two parallel sides.

Getting attraction and repulsion backwards

Parallel currents in the SAME direction attract; opposite currents repel. It is the reverse of like charges.

Magnetic Moment, Torque on a Loop, and the Galvanometer

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Magnetic Moment and the Torque on a Loop

Magnetic moment

m=NIA,τ=mBsin⁡θ,m=2πBR3μ0m = NIA, \qquad \tau = mB\sin\theta, \qquad m = \frac{2\pi BR^3}{\mu_0}

Converting a Galvanometer: the Shunt

Shunt

S=GIIg−1S = \frac{G}{\frac{I}{I_g} - 1}

Common traps

Keeping the turns fixed when a coil is rewound

The wire's length is fixed: a coil rewound to a third of the radius has three times the turns. m = NIπr² then falls to a third, not a ninth.

Using the total current over the galvanometer current

S = G/(I/I_g − 1): subtract 1, because the galvanometer still carries its own share. 4% gives G/24, not G/25.

Magnetic Field of Wires, Coils, Arcs, Solenoids and Toroids

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Field of Straight Wires

Long straight wire

B=μ0I2πdB = \frac{\mu_0 I}{2\pi d}

Field of a Circular Coil: Centre, Axis, and Rotating Charges

Circular coil

Bcentre=μ0NI2R,Baxis=μ0NIR22(R2+x2)3/2B_{\text{centre}} = \frac{\mu_0 N I}{2R}, \qquad B_{\text{axis}} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}

Arcs and Bent Wires

Arc and half-infinite wire

Barc=μ0Iθ4πR,Bhalf-infinite=μ0I4πrB_{\text{arc}} = \frac{\mu_0 I \theta}{4\pi R}, \qquad B_{\text{half-infinite}} = \frac{\mu_0 I}{4\pi r}

Solenoids, Toroids and Displacement Current

Solenoid and toroid

B=μ0nI,Btoroid=μ0NI2πrB = \mu_0 n I, \qquad B_{\text{toroid}} = \frac{\mu_0 N I}{2\pi r}

Common traps

Adding the fields of like currents at the midpoint

Midway between wires with currents in the SAME direction, their fields point opposite ways and subtract. Opposite currents add there.

Adding fields of perpendicular coils directly

Coils in perpendicular planes have perpendicular axes, so their fields add as vectors: √(B₁² + B₂²), not B₁ + B₂.

Keeping the radius when a wire is rewound

The same wire wound into n turns has a radius n times smaller. The field μ₀NI/(2R) gains n from the turns AND n from the radius: n², not n.

Counting a straight piece that points at the centre

A straight wire whose line passes through the point gives zero field there, however long it is. Only pieces that pass BESIDE the point contribute.

Adding every piece without checking its direction

Two arcs carrying current round the centre in OPPOSITE senses give opposite fields: μ₀I/4 (1/R₂ − 1/R₁), not the sum. Fix each piece's in-or-out direction before adding.

Confusing B and H in a solenoid

H = nI (A/m) is set by the winding alone; B = μ₀nI (T) includes the medium. A question asking for H in an empty solenoid wants nI, not μ₀nI.

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