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The circuit in Fig. consists of switch S, a 12.0Videal battery, a 20.0M resistor, and an air-filled capacitor. The capacitor has parallel circular plates of radius 5.00cm , separated by 3.00mm . At time t=0 , switch S is closed to begin charging the capacitor. The electric field between the plates is uniform. At t=250s, what is the magnitude of the magnetic field within the capacitor, at a radial distance 3.00cm?

Short Answer

Expert verified

The magnitude of the magnetic field within the capacitor at t=250s is, B=8.4010-13T .

Step by step solution

01

Given data

Series resistance,

R=20.0M=20.0106

Battery Voltage,=12.0V

The radius of parallel circular plate capacitor,5.00cm=5.0010-2m

Parallel plate separation distance,d=3.00cm=3.0010-2m

02

Determining the concept

To find the magnetic field in the capacitor, the current should be evaluated for a given time. For the given RC circuit current can be determined usingR, Cand, using the time constant.The capacitance depends on factors like plate area, separation distance, and permittivity of the separator. These are not normally affected by a magnetic field.

Formulae are as follows:

B=0idr2R2

i=(R)e-t/T

role="math" localid="1663220025271" C=Ad

where, B is the magnetic field,i is the current and,R is the inside radius,r is the outside radius, C is the capacitance,A is the area.

03

Determining the magnitude of the magnetic field within the capacitorfor the given circuit

Calculate the capacitance for the given dimension of the parallel plate circular capacitor and it is given as,

C=Ad

C=0蟺谤2d

C=8.8510-120.0523.0010-2

C=2.3210-11F = 23.2pF

The capacitive time constant is given as,

=RC

=20.01062.3210-11F

=4.6410-4s

At t=250s, the capacitive current is given by

i=(R)e-tT

i=1220.0106e-250.010-64.6410-4

i=3.510-5A

In capacitor,i=id, to find the magnitude of the magnetic field within the capacitor at given time t=250s,

B=0idr2R2,

whereid is the charge displacement current calculated att=250s.

Hence, the magnitude of the magnetic field B can be calculated as,

B=410-73.510-70.0320.052

B=8.4010-13T

Hence, the magnitude of the magnetic field isB=8.4010-13T.

The magnetic field for the parallel circular plate capacitor at a given time by evaluating the capacitor value, the time constant and the capacitor current can be found.

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