Complete question;
A plane circular loop of conducting wire of radius r-10 cm which possesses 15 turns is placed in a uniform magnetic field. The direction of the magnetic field makes an angle of 30° with respect to the normal direction to the loop. The magnetic field-strength is increased at a constant rate from IT to 5T in a time interval of 10 s.
a) What is the emf generated around the loop?
b) If the electrical resistance of the loop is 15Ω, what current flows around the loop as the magnetic field is increased?
Answer:
A) E.M.F generated around loop = 0.163 V
B)Current in loop; I = 0.011 A
Step-by-step explanation:
A) We are given;
Initial magnetic field strength;B1 = 1T
Final magnetic field strength;B2 = 5T
Number of turns;N = 15 turns
Radius; r = 10cm = 0.1m
Angle;θ = 30°
Time interval; Δt = 10 s
Now, the formula for magnetic flux is: Φ = NABcosθ
Where;
N is number of turns
A is area = πr²
B is magnetic field strength
θ is angle
So, initial magnetic flux is;
Φ1 = NA(B1)cosθ
Plugging in the relevant values to obtain;
Φ1 = 15*(π*0.1²)(1)cos30
Φ1 = 0.4081 Wb
Similarly, final magnetic flux is;
Φ2 = NA(B2)cosθ
Plugging in the relevant values to obtain;
Φ2 = 15*(π*0.1²)(5)cos30
Φ2 = 2.0405 Wb
The time rate of change of the flux is;
dΦ_B/dt = (Φ2 - Φ1)/Δt
So, dΦ_B/dt = (2.0405 - 0.4081)/10
dΦ_B/dt = 0.163 Wb/s
Thus, the emf generated around the loop is; E = dΦ_B/dt = 0.163 V
B) from Ohm's law, the current which flows around the loop in response to the emf is given as;
I = E/R
We are given R =15Ω
Thus; I = 0.1632/15
I = 0.011 A