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Q.(a) Define mutual inductance and write its S.I. unit.

(b) A square loop of side 'a' carrying a current I2I_2 is kept at distance xx from an infinitely long straight wire carrying a current I1I_1 as shown in the figure. Obtain the expression for the resultant force acting on the loop.
Figure — CBSE 2019 55/1/1 Q17
Figure
CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★
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(a) Mutual inductance quantifies the magnetic coupling between two circuits: the flux linked with one coil per unit current in the other. S.I. unit: henry (H). (b) The near side of the loop carries current antiparallel to the wire (repulsion) while the far side carries current parallel to it (attraction); since the near side is closer, its repulsive force wins and the net force is F=μ0I1I2a22π x(x+a)F=\dfrac{\mu_0 I_1 I_2 a^2}{2\pi\, x(x+a)}, directed away from the wire.


Part (a): Mutual Inductance

When two coils or circuits are placed near each other, a current in one produces a magnetic field that threads through the other. Mutual inductance MM captures this coupling: it is the magnetic flux linked with one coil per unit current flowing in the neighbouring coil. Equivalently, if the current in coil 1 changes at the rate dI1dt\frac{dI_1}{dt}, the emf induced in coil 2 is

E2=−MdI1dt.\mathcal{E}_2 = -M \frac{dI_1}{dt}.

The defining relation is

Φ21=MI1,\Phi_{21} = M I_1,

where Φ21\Phi_{21} is the flux through coil 2 due to current I1I_1 in coil 1.

S.I. unit: henry (H), equivalent to weber per ampere (Wb/A) or volt·second per ampere (V·s/A).


Part (b): Force on the Square Loop

Figure — CBSE 2019 55/1/1 Q17
Figure — CBSE 2019 55/1/1 Q17

Reading the figure. The infinite wire's current I1I_1 points upward. The loop's near side (the side adjacent to the wire, at perpendicular distance xx) carries I2I_2 downward — i.e. antiparallel to I1I_1. Its far side (at distance x+ax+a) carries I2I_2 upward — parallel to I1I_1. (The top and bottom sides run perpendicular to the wire and, by symmetry, contribute no net force.)

Magnetic field from the infinite wire

A long straight wire carrying current I1I_1 produces a magnetic field at perpendicular distance rr:

B(r)=μ0I12πr.B(r) = \frac{\mu_0 I_1}{2\pi r}.

Force on a current-carrying conductor

A straight segment of length ℓ\ell carrying current II in a field BB experiences a force F=BIℓsin⁡θF=BI\ell\sin\theta; for the loop's vertical sides (parallel to the wire), θ=90∘\theta=90^\circ, so F=BIℓF=BI\ell. Two long parallel currents attract if they flow in the same direction and repel if they flow in opposite directions.

Step-by-step calculation

1. Near side (distance xx, current antiparallel to I1I_1):

F1=B1I2a=μ0I1I2a2πxdirected away from the wire (repulsion).F_1 = B_1 I_2 a = \frac{\mu_0 I_1 I_2 a}{2\pi x}\quad\text{directed away from the wire (repulsion).}

2. Far side (distance x+ax+a, current parallel to I1I_1):

F2=B2I2a=μ0I1I2a2π(x+a)directed toward the wire (attraction).F_2 = B_2 I_2 a = \frac{\mu_0 I_1 I_2 a}{2\pi (x+a)}\quad\text{directed toward the wire (attraction).}

3. Top and bottom sides: each lies at a range of distances from the wire, but by symmetry the forces on these two sides are equal in magnitude and opposite in direction — they cancel exactly. …

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