CBSE Class 12 Physics NCERT Solutions: Electromagnetic Induction

Q1. A square loop of side L metres lies in the xy-plane in a region where the magnetic field is given by \($\backslash mathbf\{B\}$ = $B\_0$ (2$\backslash hat\{\backslash mathbf\{i\}\}$ + 3$\backslash hat\{\backslash mathbf\{j\}\}$ + 4$\backslash hat\{\backslash mathbf\{k\}\}$) $\backslash text\{\; T,\; where\; \}$ $B\_0$ $\backslash text\{\; is\; a\; constant.\; What\; is\; the\; magnitude\; of\; the\; magnetic\; flux\; passing\; through\; the\; square\; loop?\; \}$

• A 2B₀L² Wb
• B 3B₀L² Wb
• C 4B₀L² Wb
• D √29 B₀L² Wb

Explanation: The magnetic flux (ϕ) through a surface is given by the dot product of the magnetic field (B) and the area vector (A). For a square loop in the xy-plane, the area vector is perpendicular to the plane, i.e., along the z-axis (ϕ = L² κ̂). The magnetic field is given as \($\backslash mathbf\{B\}$ = $B\_0$ (2$\backslash hat\{\backslash mathbf\{i\}\}$ + 3$\backslash hat\{\backslash mathbf\{j\}\}$ + 4$\backslash hat\{\backslash mathbf\{k\}\}$) $\backslash text\{\; T\}$. Thus, \($\backslash phi$ = $\backslash mathbf\{B\}$ $\backslash cdot$ $\backslash mathbf\{A\}$ = $B\_0$ (2$\backslash hat\{\backslash mathbf\{i\}\}$ + 3$\backslash hat\{\backslash mathbf\{j\}\}$ + 4$\backslash hat\{\backslash mathbf\{k\}\}$) $\backslash cdot$ ($L^2$ $\backslash hat\{\backslash mathbf\{k\}\}$) = 4$B\_0$$L^2$ $\backslash text\{\; Wb\}$.}

Q2. A loop is formed by straight edges connecting points A(0,0,0), B(L,0,0), C(L,L,0), D(0,L,0), E(0,L,L), and F(0,0,L) in sequence. A magnetic field \($\backslash mathbf\{B\}$ = $B\_0$($\backslash hat\{\backslash mathbf\{i\}\}$ + $\backslash hat\{\backslash mathbf\{k\}\}$) $\backslash text\{\; T\}$ $\backslash text\{\; is\; present.\; What\; is\; the\; total\; magnetic\; flux\; passing\; through\; the\; loop\; ABCDEFA?\}$

• A B₀L² Wb
• B 2B₀L² Wb
• C √2 B₀L² Wb
• D 4B₀L² Wb

Explanation: The loop can be considered as two surfaces: the square ABCD in the xy-plane with area vector \($\backslash mathbf\{A\}$_1 = $L^2$ $\backslash hat\{\backslash mathbf\{k\}\}$\), and the square ADEFA in the yz-plane with area vector \($\backslash mathbf\{A\}$_2 = $L^2$ $\backslash hat\{\backslash mathbf\{i\}\}$\). The total area vector is \($\backslash mathbf\{A\}$ = $\backslash mathbf\{A\}$_1 + $\backslash mathbf\{A\}$_2 = $L^2$($\backslash hat\{\backslash mathbf\{i\}\}$ + $\backslash hat\{\backslash mathbf\{k\}\}$)\). The magnetic flux is \($\backslash phi$ = $\backslash mathbf\{B\}$ $\backslash cdot$ $\backslash mathbf\{A\}$ = $B\_0$($\backslash hat\{\backslash mathbf\{i\}\}$ + $\backslash hat\{\backslash mathbf\{k\}\}$) $\backslash cdot$ $L^2$($\backslash hat\{\backslash mathbf\{i\}\}$ + $\backslash hat\{\backslash mathbf\{k\}\}$) = $B\_0$$L^2$ (1 $\backslash cdot$ 1 + 1 $\backslash cdot$ 1) = 2$B\_0$$L^2$ $\backslash text\{\; Wb\}$.}

Q3. A cylindrical bar magnet is rotated about its axis. A wire connects the axis to the cylindrical surface via a contact. What happens to the current in an ammeter connected in the circuit?

• A A direct current flows.
• B No current flows.
• C An alternating sinusoidal current flows.
• D A time-varying non-sinusoidal current flows.

Explanation: When a cylindrical bar magnet rotates about its axis, the magnetic field lines passing through any stationary circuit connected to it do not change. Since the magnetic flux linked with the circuit remains constant, no electromotive force (EMF) is induced according to Faraday's Law. Consequently, no current flows through the ammeter.

Q4. Coil A moves towards a stationary coil B. A current starts flowing in B when A moves and stops when A stops moving. The current in A is counter-clockwise when observed from the side of B. What can be inferred about the current in A?

• A A constant current in the clockwise direction in A.
• B A varying current in A.
• C No current in A.
• D A constant current in the counter-clockwise direction in A.

Explanation: The fact that a current is induced in coil B only when coil A is moving, and this induced current stops when A stops moving, implies that the magnetic flux through B is changing only during the motion of A. If the current in A were varying, an EMF and current would be induced in B even if A stopped moving, due to the changing flux. Therefore, the current in A must be constant. Since the current in B stops when A stops moving, and the current in A is counter-clockwise, this implies a constant counter-clockwise current in A.