A triangular loop ABCA of resistance ‘R’ is placed at a distance ‘a’ from a long straight current carrying wire having current I. Choose the correct options.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Mutual inductance of the system in the given position is $\frac{μ_{0} a}{2 π} [1 - l n 2] .$

Mutual inductance of the system in the given position is $\frac{μ_{0} a}{2 π} [1 + l n 2]$

If the LOOP is rotated by $180^{o}$ about side AB, the charge flow in the loop is $\frac{μ_{0} i a}{2 π R} l n (\frac{27}{16})$

If the LOOP is rotated by $180^{o}$ about side AB, the charge flow in the loop is $\frac{μ_{0} i a}{π R} l n (\frac{4}{3})$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} ϕ_{i} =_{0}^{a} \frac{μ_{0} i}{2 π (a + x)} x d x \\ = \frac{μ_{0} i}{2 π}_{0}^{a} \frac{x}{(a + x)} d x \\ = \frac{μ_{0} i}{2 π}_{0}^{a} \frac{a + x - a}{(a + x)} d x \end{array}$

$\begin{array}{l} = \frac{μ_{0} i}{2 π}_{0}^{a} (1 - \frac{a}{(a + x)}) d x \\ = \frac{μ_{0} i}{2 π} {[x - a \ln (a + x)]}_{0}^{a} \\ = \frac{μ_{0} i}{2 π} [(a - a \ln (a + a)) - (0 - a \ln (a + 0))] \end{array}$

$ϕ_{i} = \frac{μ_{0} i a}{2 π} [1 - l n 2]$

$N ϕ = M i$

$M = \frac{(1) ϕ}{i} = \frac{μ_{0} a}{2 π} [1 - l n 2]$

$ϕ_{f} = \int_{0}^{a} \frac{μ_{0} i (a - x) \cos 180 °}{2 π (2 a + x)} d x = \frac{μ_{0} i}{2 π} \int_{0}^{a} \frac{(x - a)}{(2 a + x)} d x = \frac{μ_{0} i}{2 π} \int_{0}^{a} \frac{2 a + x - 2 a - a}{(2 a + x)} d x$

$\begin{array}{l} = \frac{μ_{0} i}{2 π}_{0}^{a} (1 - \frac{3 a}{(2 a + x)}) d x \\ = \frac{μ_{0} i}{2 π} {[x - 3 a \ln (2 a + x)]}_{0}^{a} \\ = \frac{μ_{0} i}{2 π} [(a - 3 a \ln (2 a + a)) - (0 - 3 a \ln (2 a + 0))] \end{array}$