Practice flux remains constant through change in magnetic flux due to a current carrying wire, and calculate the total magnetic flux through the loop due to current in the wire and through a coil.
Magnetic flux is the amount of magnetic field passing through a surface.
In other words, it is the total number of magnetic lines of field passing through a specified area in magnetic field.
It is denoted by \('\phi _m'\).
(i) Area: As the area increases, the number of field lines passing through loop increases.
Thus , \(\phi _m \propto\, A\)
(ii) Magnetic Field:
As B increases , the number of field lines passing through loop increases .
Thus , \(\phi _m \propto\, B\)
(iii) Angle between Area and Magnetic Field:
As the angle between area and magnetic field increases, the number of field lines passing through loop increases.
Thus, \(\phi _m \propto\, \) Angle between area and magnetic field
A Change in area, constant B, constant alignment
B Change in B, constant area, constant alignment
C Change in alignment, constant B, constant area
D Non - uniform but constant B, constant area, constant alignment
The magnetic flux linked with coil will be constant as the number of field lines passing through coil is constant.
Case 2 : When bar magnet is moved closer to coil.
The magnetic flux linked with coil will increase as the number of field lines passing through coil is increasing.
Case 3: When bar magnet is moved away from coil.
The magnetic flux linked with coil will decrease as the number of field lines passing through coil is decreasing.
A \(\phi _m = {\mu_0 \,I a}\, \ell n \left (1+\dfrac{a}{d}\right ) \)
B \(\phi _m =\dfrac {\mu_0 \,I a}{2\pi} \ell n \left (1+\dfrac{a}{d}\right ) \)
C \(\phi _m = {\mu_0}\,{2\pi}\, \ell n \,(I\,a) \)
D \(\phi _m =\dfrac {\mu_0 \,2\pi}{a} \ell n \left (1+\dfrac{a}{d}\right ) \)
The magnetic field and area remains constant but due to flipping, the magnetic flux through the coil changes as the angle between the area vector and magnetic field changes.
At initial position, magnetic flux is
\((\phi_m)_i = B\,A \,cos0º\)
\((\phi_m)_i = B\,A \)
\((\phi_m)_f = B\,A \,cos180º\)
\((\phi_m)_f = -B\,A \)
\(\Delta \phi_m = |(\phi_m)_f -(\phi_m)_i|\)
\(\Delta \phi_m = | -BA -BA |\)
\(\Delta \phi_m = 2\,BA \)
A \(3\, \pi \,Wb\)
B \(2\, \pi \,Wb\)
C \(5\, \pi \,Wb\)
D \(0\)
(i) Change in field with respect to time.
(ii) Change in orientation and area with respect to time of any loop or surface.
Thus, magnetic flux is constant.
(i) When loop and wire are not stationary with respect to each other.
(ii) Current in the wire is the function of time i.e., increases or decreases with respect to time.
In case of solenoid:
The change in flux through loop depends upon the current in the solenoid.
\(\vec B = B_0\,t\)
\(\phi_m = B . (Area \,of \,loop)\)
\(\phi_m = B_0 \,t \,(a^2) \,cos\,\theta\)
where,
\(\theta\) is the angle between area vector and magnetic field vector ,as shown in figure.
\(\phi_m = (B_0 \,t) a^2 \,cos\,\theta\)
A \(\ell ^2 \, B _0 \, sin \,\omega t\)
B \(\dfrac{\ell ^2}{B _0} \, sin \,\omega t\)
C \(\dfrac{\ell ^2\, B _0}{sin \,\omega t} \, \)
D \(\dfrac{ B _0 sin \,\omega \,t \, \ell _0\,}{2} \)
\(I = I _0 \,sin\,\omega t\)
\(B=\mu _0 \, n \,I\)
\(B = \mu_0 \, n \, I_0\, sin\,\omega t\)
where,
n is the number of turns per unit length .
\(\phi_m = B.A\)
\(\phi_m = BA \, cos\,0º\)
\(\phi_m = A \, \mu_0\, n \,I_0 \, sin \,\omega t\)
A \(0.12\, \pi^2 \, sin\,\omega t\, \mu Wb\)
B \(0.15\, \pi^2 \, sin\,\omega t\, \mu Wb\)
C \(0.30\, \pi^2 \, sin\,\omega t\, \mu Wb\)
D \(\pi^2 \, sin\,\omega t\, \mu Wb\)