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PhysicsEMI/ACMotional & Rotational EMFEasy2 minPYQ_2020

PhysicsEasysingle choice

A wire bent as a parabola $y = {k x}^{2}$ is located in a uniform magnetic field of induction $B$ , the vector $B$ being perpendicular to the plane $x y$ . At $t = 0$ , the sliding wire starts sliding from the vertex $O$ with a constant acceleration a linearly as shown in Fig. Find the emf induced in the loop -

Options:

Answer:

Solution:

$d ϕ = B . d A = 2 B x d y$ and $y = {k x}^{2}$

$∴ x = \sqrt{\frac{y}{k}}$

$∴ Emf E = \frac{d ϕ}{d t} = 2 B \sqrt{\frac{y}{k}} \frac{d y}{d t}$

using $v^{2} = 2 a s$

$\frac{d y}{d t} = v = \sqrt{2 a y}$

or $|ε| = \frac{d ϕ}{d t} = 2 B \sqrt{\frac{y}{k}} \sqrt{2 a y}$

or $|ε| =$ $B y \sqrt{\frac{8 a}{k}}$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Motional & Rotational EMF

⏱ 2mℹ️ Source: PYQ_2020

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