Magnetic Field Physics Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
An electron moves at 2×107 m/s2 \times 10^7 \text{ m/s} at 30°30° to a 0.1 T0.1 \text{ T} magnetic field. What is the magnetic force on the electron? (∣qe∣=1.6×10−19 C|q_e| = 1.6 \times 10^{-19} \text{ C})

Solution

  1. 1
    Use F=qvBsin⁡θF = qvB\sin\theta.
  2. 2
    F=1.6×10−19×2×107×0.1×sin⁡30°F = 1.6 \times 10^{-19} \times 2 \times 10^7 \times 0.1 \times \sin 30°
  3. 3
    F=1.6×10−19×2×107×0.1×0.5=1.6×10−13 NF = 1.6 \times 10^{-19} \times 2 \times 10^7 \times 0.1 \times 0.5 = 1.6 \times 10^{-13} \text{ N}

Answer

F=1.6×10−13 NF = 1.6 \times 10^{-13} \text{ N}
When the velocity is not perpendicular to the field, only the perpendicular component contributes to the magnetic force. The parallel component causes the charge to spiral.

About Magnetic Field

A vector field around magnets and moving charges that exerts force on other moving charges and magnetic materials.

Learn more about Magnetic Field →

More Magnetic Field Examples

Example 1 medium

A proton ([formula]) moves at [formula] perpendicular to a magnetic field of [formula]. What force d

Example 2 hard

A wire carrying [formula] of current is [formula] long and placed perpendicular to a [formula] magne

Example 4 hard

A proton ([formula], [formula]) enters a uniform magnetic field of [formula] at [formula] with a spe

← All Magnetic Field Examples Magnetic Field Concept