WEBVTT 00:00:00.066 --> 00:00:02.733 This is College Physics Answers with Shaun Dychko. 00:00:02.980 --> 00:00:05.393 The force due to the magnetic field on the charge is 00:00:05.473 --> 00:00:08.620 the charge times its speed times the magnetic field strength. 00:00:08.913 --> 00:00:12.100 This is assuming the charge velocity is perpendicular to the magnetic field 00:00:12.100 --> 00:00:13.446 which it is, in this question. 00:00:14.233 --> 00:00:17.273 And that magnetic force is going to be providing 00:00:17.273 --> 00:00:20.333 the centripetal force of the charge needs to go in a circle. 00:00:20.546 --> 00:00:23.246 And so the centripetal force is the mass of the charge 00:00:23.246 --> 00:00:26.466 times its speed squared divided by the radius of the circle. 00:00:27.446 --> 00:00:31.520 So we can solve this for B by dividing both sides by qv. 00:00:33.800 --> 00:00:38.266 And we get that the magnetic field is mv squared over qvr. 00:00:38.266 --> 00:00:41.913 And so one of the v is cancel and we end up with mv over qr. 00:00:43.420 --> 00:00:46.273 So the magnetic field strength is the mass of the proton 00:00:46.273 --> 00:00:50.080 which is 1.67 times ten to the minus 27 kilograms times the speed 00:00:50.080 --> 00:00:52.433 which is 7.5 times ten to the seven meters per second 00:00:52.780 --> 00:00:55.433 divided by the charge on the proton which is the elementary charge 00:00:55.433 --> 00:00:59.140 1.6 times ten to the minus 19 Coulombs times the radius of the path 00:00:59.140 --> 00:01:03.633 which we're told is 0.8 meters, giving us magnetic field strength of 0.979 Tesla.