Part A A particle of mass 1.2782 × 1025 kg and charge of 3.2×1019 C is accelerat

Question

Part A A particle of mass 1.2782 × 1025 kg and charge of 3.2×1019 C is accelerated from rest in the plane of the page through a potential difference of 454 V between two parallel plates as shown. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0.077 T. The particle curves in a semicircular path and strikes a detector. What is the magnitude of the force exerted on the charged particle as it enters the region of the magnetic field B ? Answer in units of N.

Part B

A circular coil consisting of a single loop of wire has a radius of 18.3 cm and carries a current of 10.2 A. It is placed in an external magnetic field of 0.338 T.

Find the magnitude of the torque on the wire when the plane of the coil makes an angle of 15.2 with the direction of the field.

Answer in units of N · m.

Explanation / Answer

change in kinetic energy = q deltaV

1.2782 x 10^-25 v^2 / 2 = (3.2 x 10^-19) (454)

v = 47678 m/s

F = q v B = (3.2 x 10^-19) (47678) (0.077)

F = 1.175 x 10^-15 N ...........Ans

(B) u = I A = pi x 0.183^2 x 10.2 = 1.073 A m^2

B = 0.338 T

torque = u x B

= u B sin(theta)

= (1.072) (0.338) sin(90 - 15.2)

= 0.35 N m

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