E16D.1 Consider a circular loop of radius R that is expand- ing at a rate of dR/dt in a static and uniform magnetic field B that is perpendicular to the loop's face. Since the mag- netic fiel...
E16D.1 Consider a circular loop of radius R that is expand- ing at a rate of dR/dt in a static and uniform magnetic field B that is perpendicular to the loop's face. Since the mag- netic field is static and uniform in every frame of reference, Faraday's law ХЕ + дв/at 0 tells us that the electric field in any frame in this situation must have zero curl and therefore cannot drive a current around the loop. (a) At a certain instant of time, divide the loop into a set of counterclockwise steps dř, calculate the work done by the magnetic field on a positive charge carrier moving through each step, and sum to find the total magnetic emf (energy per unit charge) that would be delivered to a hypothetical charge carrier going instantaneously around the loop. (This is actuall what a voltmeter attached to the loop would measure at that instant.) (b) Show that even though Faraday's law does not apply to this situation, the emf calculated in part (a) is cor- rectly given by Faraday's law of induction.
E16D.1 Consider a circular loop of radius R that is expand- ing at a rate of dR/dt in a static and uniform magnetic field B that is perpendicular to the loop's face. Since the mag- netic field is static and uniform in every frame of reference, Faraday's law ХЕ + дв/at 0 tells us that the electric field in any frame in this situation must have zero curl and therefore cannot drive a current around the loop. (a) At a certain instant of time, divide the loop into a set of counterclockwise steps dř, calculate the work done by the magnetic field on a positive charge carrier moving through each step, and sum to find the total magnetic emf (energy per unit charge) that would be delivered to a hypothetical charge carrier going instantaneously around the loop. (This is actuall what a voltmeter attached to the loop would measure at that instant.) (b) Show that even though Faraday's law does not apply to this situation, the emf calculated in part (a) is cor- rectly given by Faraday's law of induction.