# Electrical Engineering - Magnetism and Electromagnetism - Discussion

Discussion Forum : Magnetism and Electromagnetism - General Questions (Q.No. 2)
2.
The induced voltage across a coil with 250 turns that is located in a magnetic field that is changing at a rate of 8 Wb/s is
1,000 V
2,000 V
31.25 V
3,125 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Dodo said:   7 years ago
The induced voltage is a multiplication of a number of turns and rate of change of flux.

Hence,

V = Ndphi/dt,

V = 250*8.
V = 2000V.
(1)

Mukesh kumar pandey said:   7 years ago
Induced voltage is equal than multiple for the rate of change of flux into number of turns I.e v = 250 * 8 = 2000 volt.

Mahadev nagargoje said:   8 years ago
The induced voltage is a multiplication of a number of turns and rate of change of flux.

Rathod prem said:   8 years ago
Let us consider;

Here V = N * {d(φ) /dt}.
N = No of turns of coil = 250 turns.

D(φ) /dt = Rate of change of flux = 8wb/sec.

Hence V = 250 * 8 = 2000.
(3)

Amita said:   8 years ago
Induced voltage is a multiplication of a number of turns and rate of change of flux.

Azaria said:   9 years ago
250*8=2000 v.

Venkatesh said:   10 years ago
Number of turns multiplied by rate of change of flux linkages gives the induced voltage in the machine.

Faheem said:   1 decade ago
Induced voltage= - no of turns x rate of change of mag.flux.

Shobhan Kumar Dash said:   1 decade ago
V= N* {d(phi)/dt}

Here N = No. of turns of coil = 250 turns
d(phi)/dt = Rate of change of flux = 8 Wb/Sec.

Hence V= 250 * 8 = 2000 Volt (Ans)

Bk samal said:   1 decade ago
e=speed*rate of change of flux

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