The figure shows a square loop L of side 5, c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $1$. Analyze the resistance network: The network is a Wheatstone bridge with arms $1\, \Omega, 2\, \Omega, 1\, \Omega, 2\, \Omega$ and a central $

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) $1$. Analyze the resistance network: The network is a Wheatstone bridge with arms $1\, \Omega, 2\, \Omega, 1\, \Omega, 2\, \Omega$ and a central $3\, \Omega$ resistor. Since $\frac{1}{1} = \frac{2}{2}$,it is a balanced Wheatstone bridge. The central $3\, \Omega$ resistor carries no current. The equivalent resistance of the bridge is $R_{eq} = \frac{(1+2)(1+2)}{(1+2)+(1+2)} = \frac{3 	imes 3}{3+3} = 1.5\, \Omega$.$2$. Calculate the induced electromotive force $(EMF)$: The motional $EMF$ is given by $\varepsilon = B \ell v$. Here,$B = 1\, T$,$\ell = 5\, cm = 0.05\, m$,and $v = 1\, cm/s = 0.01\, m/s$. Thus,$\varepsilon = 1 	imes 0.05 	imes 0.01 = 5 	imes 10^{-4}\, V$.$3$. Calculate the total resistance: The total resistance $R_{total} = R_{loop} + R_{eq} = 1.7\, \Omega + 1.5\, \Omega = 3.2\, \Omega$.$4$. Calculate the current: The current $i = \frac{\varepsilon}{R_{total}} = \frac{5 	imes 10^{-4}}{3.2} \approx 1.5625 	imes 10^{-4}\, A = 156.25\, \mu A$. The closest option is $150\, \mu A$ or $170\, \mu A$. Re-evaluating the bridge: $R_{eq} = \frac{3 	imes 3}{3+3} = 1.5\, \Omega$. Total $R = 1.7 + 1.5 = 3.2\, \Omega$. $i = 156.25\, \mu A$. Given the options,$150\, \mu A$ is the closest value.

Similar Questions

The migratory pattern of birds is one of the mysteries in the field of biology and indeed all of science. Explain this in the context of electromagnetic induction.

$A$ closed loop is moving with constant velocity in a uniform magnetic field as shown. The potential difference and current in element $AB$ will be:

$A$ straight conductor of length $0.4 \ m$ is moving with a speed of $7 \ ms^{-1}$ perpendicular to a magnetic field of intensity $0.9 \ Wb \ m^{-2}$. The induced emf across the conductor will be (in $V$)

$A$ jet plane is travelling towards west at a speed of $1800\; km/h$. What is the voltage difference developed between the ends of the wing having a span of $25\; m$,if the Earth's magnetic field at the location has a magnitude of $5 \times 10^{-4}\; T$ and the dip angle is $30^{\circ}$ (in $; V$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module