A conducting square frame of side ‘a’ and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ‘V’. The emf induced in the frame will be proportional to

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{1}{(2 x + a)^{2}}$

$\frac{1}{x^{2}}$

$\frac{1}{(2 x - a) (2 x + a)}$

$\frac{1}{(2 x - a)^{2}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Potential difference across AD
Question Image
$V_{D} - V_{A} = Bav = \frac{μ_{0} iav}{2 π (x - \frac{a}{2})}$
Potential difference across= BC
$\begin{array}{l} V_{C} - V_{B} = Bav = \frac{μ_{0} iav}{2 π (x + \frac{a}{2})} \\ V_{net} = (V_{D} - V_{A}) - (V_{C} - V_{B}) \\ = \frac{μ_{0} iav}{2 π} [\frac{a}{(x - \frac{a}{2}) (x + \frac{a}{2})}] = \frac{μ_{0} {ia}^{2} v}{π (2 x - a) (2 x + a)} \\ ∴ V_{net} \propto \frac{1}{(2 x - a) (2 x + a)} \end{array}$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET