To solve Problem 26, let’s analyze the situation carefully and provide a step-by-step solution.

Given:

• A charged particle moves through a uniform magnetic field $B$, follows a half-circle, and exits.
• Time spent in the field: $t = 130 \ \text{ns} = 130 \times 10^{-9} \ \text{s}$.
• The particle is either a proton or an electron (we must decide which one).
• In part (b), the kinetic energy is doubled, and we are to determine the new time spent in the field.

Part (a): Find the magnitude of the magnetic field $B$

We know that the motion of the charged particle in a magnetic field follows a circular path due to the Lorentz force. The time taken to complete a full circle $T$ is related to the charge-to-mass ratio, magnetic field, and velocity:

$T = \frac{2\pi m}{qB}$

• For half a circle, the particle takes half the period $T$. Thus:

$t_{\text{half}} = \frac{T}{2} = \frac{\pi m}{qB}$

Rearranging for $B$, we get:

$B = \frac{\pi m}{q t_{\text{half}}}$

Here:

• $t_{\text{half}} = 130 \ \text{ns} = 130 \times 10^{-9} \ \text{s}$,
• For a proton:
  • Charge $q = 1.6 \times 10^{-19} \ \text{C}$,
  • Mass $m = 1.67 \times 10^{-27} \ \text{kg}$.

Substitute the values:

$B = \frac{\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19}) (130 \times 10^{-9})}$

Simplify step by step:

1. Compute the numerator:

$\pi \times 1.67 \times 10^{-27} \approx 5.24 \times 10^{-27}$

2. Compute the denominator:

$(1.6 \times 10^{-19}) (130 \times 10^{-9}) = 2.08 \times 10^{-26}$

3. Divide:

$B = \frac{5.24 \times 10^{-27}}{2.08 \times 10^{-26}} \approx 0.252 \ \text{T}$

Thus, the magnitude of the magnetic field is:

$\boxed{B \approx 0.25 \ \text{T}}$

Part (b): Time for the second trip (Kinetic energy doubled)

If the kinetic energy of the particle is doubled, the velocity increases. Kinetic energy is proportional to the square of the velocity:

$K \propto v^2$

If $K' = 2K$, then:

$v' = \sqrt{2} v$

Now, the time to complete a circular path is:

$T = \frac{2\pi r}{v}$

In a magnetic field, the radius $r$ of the circular path is proportional to the velocity:

$r = \frac{m v}{q B}$

Thus, the time for one full circle becomes:

$T = \frac{2\pi r}{v} = \frac{2\pi \left( \frac{m v}{q B} \right)}{v} = \frac{2\pi m}{q B}$

We see that the period $T$ does not depend on the velocity, so doubling the kinetic energy does not change the period.

Therefore, the time for the half-circle remains the same as before:

[ t_{\text{half}} = 130 \ \text{ns} \div \sqrt{2} =