This question is incomplete, the complete question is;
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.
At what rate is the magnetic field changing?
Answer:
the magnetic field changing at the rate of 9.33 m T/s
Explanation:
Given the data in the question;
Electric field E = 7 mV/m
radius r = 1.5 m
Now, from Faraday law of induction;
∫E.dl = d∅/dt
E∫dl = A( dB/dt )
E( 2πr ) = πr² ( dB/dt )
( 0.007 ) = (r/2) ( dB/dt )
( 0.007 ) = 0.75 ( dB/dt )
dB/dt = 0.007 / 0.75
dB/dt = 0.00933 T/s
dB/dt = ( 0.00933 × 1000) m T/s
dB/dt = 9.33 m T/s
Therefore, the magnetic field changing at the rate of 9.33 m T/s
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution
Answer:
The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.
F = N e E where E is the value of the field and N e the charge Q
M g = N e E and M g is the weight of the drop
N = M g / (e E)
N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13
.00011 kg is a very large drop
Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs
Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons
Speed = distance / time
3.4cm / 0.1s = 34 cm/sec
