Chapter 26.1, Problem 3PP

To determine

The speed of the moving protons if the magnetic field is 0.060 T and electric field is $9.0\times {10}^3\text{V}\cdot {\text{m}}^{\text{-1}}$ .

Answer to Problem 3PP

The speed of the electrons is $\boxed{v=15\times {10}^3{\text{ms}}^{\text{-1}}}$ .

Explanation of Solution

Given:

The magnitude of electric field is $E=9.0\times {10}^3\text{V}\cdot {\text{m}}^{\text{-1}}$ .

Strength of magnetic field is $B=0.060\text{T}$

Formula used:

The magnetic force $F_B$ is expressed as

  $F_B=Bqv$  ...... (1)

The electric force $F_E$ is expressed as

  $F_E=Eq$  ...... (2)

Here, $B,E,q$ are the strength of magnetic field, magnitude of electric field and

Calculation:

The magnetic field $F_B$ experienced by a charged particle gets balanced with the electrostatic force $F_E$ experienced by it in the equilibrium position, so

  $F_B=F_E$  ...... (3)

Substitute $Bqv$ for $F_B$ and $Eq$ for $F_E$ in equation (3)

  $Bqv=Eq$

  $v=\frac{E}{B}$  ...... (4)

Substitute 0.60 T for $B$ and $9\times {10}^3\text{N}\cdot {\text{C}}^{\text{-1}}$ for $E$ in equation (4)

  $v=\frac{9\times {10}^3\text{N}\cdot {\text{C}}^{\text{-1}}}{0.60\text{T}}$

  $\boxed{v=15\times {10}^3{\text{ms}}^{\text{-1}}}$

Conclusion:

Hence, the speed of the electrons is $\boxed{v=15\times {10}^3{\text{ms}}^{\text{-1}}}$ .