The force exerted by a magnetic field on a wire carrying current is:
where I is the current, L the length of the wire, B the magnetic field intensity, and
the angle between the wire and the direction of B.
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is
, so we can re-arrange the formula and substitute the numbers to find B:
Explanation:
M₂ = Fr²/GM₁
M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]
M₂ = (7.79N*m²)/(7.84*10^-7N*m²)
M₂ = 9.94*10^6 kg
Answer:
1 × 10⁶ N/C
Explanation:
The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C
Answer:
0.5 A
Explanation:
N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm
The induced emf is given by
e = - N dФ/dt
Where, dФ/dt is the rate of change of magnetic flux.
Ф = B A
dФ/dt = A dB/dt
so,
e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V
negative sign shows the direction of magnetic field.
induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g
= 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = 
⇒ T = 4
M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × 
×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
