<em>Answer</em>
0.6 teslas
<em>Explanation</em>
When a conductor is inside a magnetic field it experiences a force given by;
Force = ILBsinθ
Where I⇒ current
L ⇒length of the conductor
B ⇒ magnetic field strength
θ ⇒ Angle between the conductor and magnetic field.
F = ILBsinθ
When θ = 90°, Then sin 90 =1 and the formula becomes;
F =ILB
3 = 10 × 0.5 × B
3 = 5B
B = 3/5
= 0.6
magnetic field strength = 0.6 teslas
The current in the second diode is 400mA
Data;
- First Voltage = 0.75V
- Second Voltage = 0.8V
- First Current (I) = 400mA
- Second Current(I) = ?
<h3>Current In a Series</h3>
The current in the first diode is equal to 400mA. In a series circuit, the current passing the diodes are equal. This implies that the current in the series are equal.
Diodes connected in series will be the equal.
Since I1 is 400mA, I2 will be equal to 400mA
The current in the second diode is 400mA
Learn more on current in a diode here;
brainly.com/question/1455378
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brainly.com/sf/question/2641701
Answer: the electric field is equal to: E=0 for r<2 cm; E= -2.29*10^4/r^2 for 2cm<r<4cm; E=0 for 4cm<r<5 cm; E=5.26*10^4/r^2 for r>5 cm
the electric field in N/C units
Explanation: In order to find the electric field for all r values, we have to use the definition of electric field and Gaussian law.
In this sense, for r<2 cm as it is inside a conductor teh electric field is zero.
for 2cm< r< 4 cm we applied the field from a spherical charge distribution so by the Gaussian law we find the total charge inside the gaussian surface so
E.4π r^2= Q inside/ε0 = -2.55μC/ε0
Idem for other regions.
for 4 cm<r< 5 cm in the outsider conductor the E=0
Finally, for r>5 cm
E.4π r^2=Q inside/ε0=(8.40-2.55)μC/ε0
