Question

"An electron is moving at 3.0 × 106 m/s perpendicular

to a uniform magnetic field. If the radius of the motion is 18 mm,
what is the magnitude of the magnetic field?

Answer 1

The magnitude of the magnetic field is approximately 0.333 T.

Step-by-step explanation:

When an electron moves perpendicular to a uniform magnetic field, it experiences a centripetal force due to the magnetic force acting on it. The centripetal force is provided by the magnetic force, given by the equation
`\(F = (mv^2)/(r)\)`.

The magnetic force is also given by the equation
`\(F = qvB\)`. Setting these two expressions for
`\(F\)`equal to each other, we get
`\((mv^2)/(r) = qvB\).`

Solving for
`\(B\), we find \(B = (mv)/(qr)\).` Substituting the given values
`(mass of an electron \(m \approx 9.109 * 10^(-31)\)`kg, charge of an electron
`\(q \approx -1.602 * 10^(-19)\) C, velocity \(v = 3.0 * 10^6\) m/s, and radius \(r = 18 * 10^(-3)\) m),`we calculate
`\(B\) to be approximately 0.333 T.`

Therefore, the magnitude of the magnetic field is approximately 0.333 T. This calculation demonstrates the relationship between the centripetal force experienced by the electron and the magnetic force acting on it in a uniform magnetic field.

Answer 2

To calculate the magnitude of the magnetic field, we can use the formula: r = mv / (qB). Plugging in the given values, B ≈
`6.88 * 10^{-4 T`.

Step-by-step explanation:

To calculate the magnitude of the magnetic field, we can use the formula:

r = mv / (qB)

Where:

• r is the radius of the motion
• m is the mass of the electron
• v is the velocity of the electron
• q is the charge of the electron
• B is the magnetic field strength

Plugging in the given values:

`18 mm = (9.11 * 10^(-31 )kg) * (3.0 * 10^6 m/s) / (1.6 * 10^(-19) C) * B`

Solving for B:

`B = ((9.11 x 10^(-31) kg) * (3.0 * 10^6 m/s)) / ((1.6 * 10^(-19) C) * (18 * 10^(-3) m))`

B ≈ 6.88 x
`10^(-4)` T