The emf induced in the coil is -5.65 V
<h3>Induced emf in coil</h3>
The induced emf in the coil is given by ε = -NΔΦ/Δt where
- ΔΦ = change in magnetic flux Φ₂ - Φ₁ where
- Φ₁ = initial magnetic flux = -58 Wb and
- Φ₂ = final magnetic flux = 38 Wb and and
- Δt = change in time = t₂ - t₁ where
- t₁ = initial time = 0 s and
- t₂ = final time = 34 sand
- N = number of loops of coil = 2
Since ε = -NΔΦ/Δt
ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)
Substituting the values of the variables into the equation, we have
ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)
ε = -2(38 Wb - (-58 Wb))/(34 s - 0 s)
ε = -2(38 Wb + 58 Wb)/(34 s - 0 s)
ε = -2(96 Wb)/34 s
ε = -192 Wb/34 s
ε = -5.65 Wb/s
ε = -5.65 V
So, the emf induced in the coil is -5.65 V
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Answer:
Explained
Explanation:
1.Each of the spring scale will read 10N,considering acceleration due to gravity as 10 m/s^2
2.Each of the spring scale will read 10N because each string exerts a force of 10 N to counterbalance the force of 1 kg mass attached to it. This means the tension on the both side of the string is 10 N. So the scale will read 10 N. Also as spring balances are attached in series and kept on table so both spring balances will read same readings.
Answer:
Explanation is in the picture and the answer is 16
Answer:
Explanation:
Let m and e are the mass and charge of an electron. It is accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. Let v is the velocity of the electron. It can be calculated as :
When the electron enters the magnetic field, the centripetal force is balanced by the magnetic force as :
or
So, the radius of the resulting electron trajectory is 
. Hence, this is the required solution.
