SATHEE: Electro Magnetic Induction And Alternating Currents Question 448

Electro Magnetic Induction And Alternating Currents Question 448

Question: A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. The emf induced across the rod is

Options:

A) $\frac{μ_{0} i v^{2}}{2 π} l n (\frac{r + L}{R})$

B) $\vec{B}$

C) $\frac{μ_{0} i v}{2 π} l n (\frac{r + L}{R})$

D) $\frac{μ_{0} i v}{2 π} l n (\frac{r^{2} + L^{2}}{L^{2}})$

Show Answer

Answer:

Correct Answer: C

Solution:

Consider a small element of length dx of the rod at a distance x and (x+dx) from the wire. The emf induced across the element

$d e = B v d x$ -.(i)

We know that magnetic field B at a distance x from a wire carrying a current; is given by

$B = \frac{μ_{0}}{2 π} . \frac{i}{x}$ .. (ii)

From eqs. (i) and (ii),

$d e = \frac{μ_{0} i}{2 π x} v d x$ …(iii) The emf induced in the entire length of the rod PQ is given by

$e = \int d e = \int_{P}^{Q} \frac{μ_{0}}{2 π} \frac{i}{x} v d x$

$= \int_{r}^{r + L} \frac{μ_{0}}{2 π} \frac{i}{x} v d x = \frac{μ_{0}}{2 π} i v \int_{r}^{r + L} \frac{d x}{x}$

$= \frac{μ_{0} i v}{2 π} \log_{e} (\frac{r + L}{r})$

Electro Magnetic Induction And Alternating Currents Question 448

Question: A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. The emf induced across the rod is

Options:

Answer:

Solution:

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Electro Magnetic Induction And Alternating Currents Question 448

Question: A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. The emf induced across the rod is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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