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Fields - Magnetic Fields

Grade 12IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A magnetic field is a region of space where a magnetic pole, a moving charge, or a current-carrying conductor experiences a force.

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The magnetic field strength BB (also called magnetic flux density) is measured in Tesla (TT), where 1T=1NA−1m−11 T = 1 N A^{-1} m^{-1}.

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Magnetic field lines point from North to South outside a magnet. The density of lines represents the strength of the field.

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Magnetic force on a moving charge: F=qvBsin⁡θF = qvB \sin\theta. The force is always perpendicular to both the velocity vv and the magnetic field BB.

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The direction of the force on a positive charge is given by the Right-Hand Rule (or Fleming's Left Hand Rule). For a negative charge, the force direction is reversed.

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Motion in a uniform field: Since the force is always perpendicular to the velocity, a charged particle will undergo uniform circular motion with a centripetal force Fc=mv2r=qvBF_c = \frac{mv^2}{r} = qvB.

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Magnetic force on a current-carrying conductor: F=BILsin⁡θF = BIL \sin\theta, where II is the current and LL is the length of the conductor within the field.

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Force between two parallel current-carrying wires: If currents are in the same direction, they attract; if they are in opposite directions, they repel.

📐Formulae

F=qvBsin⁡θF = qvB \sin\theta

F=BILsin⁡θF = BIL \sin\theta

r=mvqBr = \frac{mv}{qB}

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

FL=μ0I1I22πr\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}

💡Examples

Problem 1:

An electron (q=−1.6×10−19Cq = -1.6 \times 10^{-19} C, m=9.11×10−31kgm = 9.11 \times 10^{-31} kg) enters a uniform magnetic field of 0.50T0.50 T perpendicularly with a speed of 4.0×106ms−14.0 \times 10^6 m s^{-1}. Calculate the radius of the electron's circular path.

Solution:

Using the formula for the radius of a charged particle in a magnetic field: r=mvqBr = \frac{mv}{qB} Substituting the values: r=(9.11×10−31kg)(4.0×106ms−1)(1.6×10−19C)(0.50T)r = \frac{(9.11 \times 10^{-31} kg)(4.0 \times 10^6 m s^{-1})}{(1.6 \times 10^{-19} C)(0.50 T)} r≈4.56×10−5mr \approx 4.56 \times 10^{-5} m

Explanation:

Because the electron enters the field perpendicularly (sin90∘=1\\sin 90^\circ = 1), the magnetic force acts as the centripetal force. Setting qvB=mv2rqvB = \frac{mv^2}{r} allows us to solve for rr.

Problem 2:

A straight wire of length 0.20m0.20 m carries a current of 5.0A5.0 A in a direction that makes an angle of 30∘30^\circ with a uniform magnetic field of 0.15T0.15 T. Calculate the magnitude of the magnetic force on the wire.

Solution:

Using the formula: F=BILsin⁡θF = BIL \sin\theta Substituting the values: F=(0.15T)(5.0A)(0.20m)sin⁡(30∘)F = (0.15 T)(5.0 A)(0.20 m) \sin(30^\circ) F=(0.15)(5.0)(0.20)(0.5)F = (0.15)(5.0)(0.20)(0.5) F=0.075NF = 0.075 N

Explanation:

The force on a current-carrying wire depends on the component of the magnetic field perpendicular to the current. The sin⁡θ\sin\theta term accounts for the angle between the wire and the field lines.

Magnetic Fields - Revision Notes & Key Formulas | IB Grade 12 Physics