Name: Magnetic Flux
Uploaded: 2019-07-19
Duration: 6 min 44 s
Description: Magnetic flux is the total number of magnetic lines of force that are passing through a given surface. To learn more about this intriguing concept, watch the...

Question

Consider the mutual inductance between a long straight wire and a square loop of side $a$ as shown in the figure below.
Now assume that the straight wire carries a current of $50 A$ and the loop is moved to the right with a constant velocity, $v = 10 m / s$ . Calculate the induced emf in the loop at the instant when $x = 0.2 m$ . Take $a = 0.1 m$ and assume that the loop has a large resistance.

Accepted Answer

Given,
The wire is carrying a current $I = 50 A$
Velocity of the loop $v = 10 m / s$

Side of the loop $a = 0 .1 m$

Distance of the closer side of loop $x = 0 .2 m$

The respective emfs will be induced in the sides that are parallel to the infinite wire while it will be zero in the sides which are perpendicular to the wire.

When the closer side of loop is at a distance $x$ from the current carrying wire:
Induced emf in the closer side of loop is, $ε_{1} = B_{1} v a = \frac{μ_{0} I}{2 π x} v a$
where $B_{1}$ is the field at a distance $x$ .
Induced emf in the farther side of loop is, $ε_{2} = B_{2} v a = \frac{μ_{0} I}{2 π (x + a)} v a$
where $B_{2}$ is the field at a distance $x + a$ .
Net Induced emf, $ε = ε_{1} - ε_{2} = \frac{μ_{0} I}{2 π x} v a - \frac{μ_{0} I}{2 π (x + a)} v a = \frac{μ_{0} I}{2 π} v a (\frac{1}{x} - \frac{1}{x + a})$
$\Rightarrow ε = 2 \times 10^{- 7} \times 50 \times 10 \times 0.1 \times (\frac{1}{0.2} - \frac{1}{0.2 + 0.1})$
$\Rightarrow ε = 1.7 \times 10^{- 5} V$

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