Magnetic Field Help

1cm

The isotopes are singly ionized, which means that a single electron has been removed from each atom. And the isotopes are now having charge of e = 1.60 10 19 C . 6 From the relation rqB mv mv 2 , and thus r = . = qvB , we obtain v = qB m r r235 = r238 = mv (3.90 10 25 kg )(1.05 10 5 m / s ) = 34.1 cm . = qB (1.60 10 19 C )(0.750 T ) mv (3.95 10 25 kg )(1.05 10 5 m / s ) = 34.6 cm . = qB (1.60 10 19 C )(0.750 T ) The separation between the isotopes: d = 2r238 2r235 = 2(34.6 cm 34.1 cm) = 1 cm . 16.4 The magnetic force exerted on a current-carrying wire As a charged particle experiences a force when it moves across magnetic field lines. The same thing happens when a wire carries current on it. Consider a straight segment of length L of a wire with a current I flowing from left to right, presents in a magnetic field B, as shown in figure. If the conducting charges move through the wire with an average speed v, the time required for them to move from one end of the wire segment to the other is t = L / v . The amount of charge that flows through the wire in this time is q = I t = IL / v . Therefore, the force exerted on the wire is I L F = qvB sin = vB sin . v Hence, we have F = ILB sin . Maximum force occurs when the current is perpendicular to the magnetic field ( = 90o) and is zero if the current is in the same direction as B ( = 0o). The direction of the magnetic force is given by the RHR, where the direction of charge velocity v is now the direction of current I.

Thanks guys, good explanation as well guest. Saved me on my test Wednesday!