Final answer:
The value of q/m for a particle in a 0.53 T magnetic field, whose path is made straight by a crossed 280 V/m electric field, is approximately 130845.07 C/kg. This value is obtained by balancing the electric and magnetic forces and then solving for q/m.
Step-by-step explanation:
The question asks what is the value of q/m for a particle that moves in a circle of radius 7.2 mm in a 0.53 T magnetic field if a crossed 280 V/m electric field will make the path straight. To make the particle's path straight, the electric force (FE) must balance the magnetic force (FB). The electric force is given by FE = qE, and the magnetic force is given by FB = qvB, where q is the charge, v is the velocity, and B is the magnetic field strength.
If the electric and magnetic forces are balanced, then qE = qvB. Since we want to find q/m, we can divide both sides by m and by B to get q/m = E/vB. The radius r given by the centripetal force needed for circular motion is also given by the magnetic force: mv2/r = qvB, so we can rearrange this to solve for velocity v = qBr/m. Substituting back this into the previous equation for q/m we get q/m = E/(B2r). Plugging in the values E = 280 V/m, B = 0.53 T, and r = 7.2 mm = 0.0072 m, we find the charge-to-mass ratio q/m.
To calculate the actual value: q/m = 280 V/m / (0.53 T)2 x 0.0072 m) = 280 / 0.532 x 0.0072 = 130845.07 C/kg (approximately).