Answer:
-1.627×10⁻³ V.
Explanation:
Applying,
E = -dФ/dt................ Equation 1
Where E = induced emf, dФ = change in magnetic flux, dt = change in time.
But,
Ф = BA
Where, B = magnetic Field, A = Area of the loop
Since the loop is a square,
A = l²
Where l = length of the loop
Therefore,
Ф = Bl²........................ Equation 2
Substitute equation 2 into equation 1
E = -d(Bl²)/dt
E = -2Bldl/dt.............. Equation 3
Given: B = 0.35 T, l = 8.3 cm = 0.083 m, dl/dt = 2.8 cm/s = 0.028 m/s.
Substitute into equation 3
E = -2(0.35×0.083×0.028)
E = -1.627×10⁻³ V.
Hence the induced emf = -1.627×10⁻³ V
Note: The negative sign shows that the induced emf opposes the motion that produces it.
Answer:
The induced emf in the loop is 1.627 mV
Explanation:
Given;
magnetic field strength, B = 0.35 T
the length of each side of the square is decreasing at a constant rate, dl/dt = 2.8 x 10⁻² m/s
Induced emf is given as;
[tex]emf = -\frac{d \phi}{dt}[/tex]
where;
Ф is magnetic flux through the loop, given as;
Ф = BA
A is area, area of square = L²
Ф = BL²
[tex]emf =- \frac{d \phi}{dt} = -\frac{d(Bl^2)}{dt} \\\\emf = -2Bl\frac{dl}{dt}[/tex]
Substitute in the given values;
[tex]emf = -2(0.35*0.083)(-2.8*10^{-2}) = 1.627 *10^{-3} \ V\\\\emf = 1.627 \ mV[/tex]
Therefore, the induced emf in the loop is 1.627 mV
