Chapter 27, Problem 50P

The magnetic field inside a solenoid of circular cross section is given by $\stackrel{\to }{B}$ = bt $\hat{k}$, where b = 2.1 T/ms. At time t = 0.40 μs, a proton is inside the solenoid at x = 5.0 cm, y = z = 0, and is moving with velocity $\stackrel{\to }{v}$ = 4.8ĵ Mm/s. Find the electromagnetic force on the proton.

50. Interpret The problem involves a changing magnetic field that induces an electric field. Both fields exert a force on a proton.

Develop The induced electric field inside a long thin cylindrical solenoid, whose magnetic field is increasing with time as $\stackrel{\to }{B}$ = bt $\hat{k}$. (take $\hat{k}$ into the page as in Fig. 27.30). can be found by modifying the argument in Example 27.11 for radius r < R. The magnitude of $\stackrel{\to }{E}$ is

$E=\left|-\frac{1}{2}r\frac{dB}{dt}\right|=\frac{rb}{2}$

The induced electric field circulates CCW around the direction of $\stackrel{\to }{B}$ or $\stackrel{\to }{E}$ = −(bx/2) ĵ for the given point on the x axis (x = 5 cm, y = z = 0) inside the solenoid (whose axis we assume is the z axis). At t = 0.4 μs, the uniform magnetic field is $\stackrel{\to }{B}$ = b(0.4 μs) $\hat{k}$. The electromagnetic force on a proton with the given velocity is $\stackrel{\to }{F}$ = e( $\stackrel{\to }{E}$ + $\stackrel{\to }{v}$ × $\stackrel{\to }{B}$).

Evaluate Substituting the values given, we find the net force on the proton to be