A rectangular loop of length 2.5 m and width | vedclass

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Question

(C) The magnetic flux $\phi$ through the loop is given by $\phi = B A \cos 	heta$,where $	heta$ is the angle between the magnetic field vector and t

Vedclass · Accepted Answer

$+1 \ V$

Vedclass · Answer

(C) The magnetic flux $\phi$ through the loop is given by $\phi = B A \cos 	heta$,where $	heta$ is the angle between the magnetic field vector and the area vector.Given,the loop is placed at $60^{\circ}$ to the magnetic field,so the angle between the area vector and the magnetic field is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.Initial flux $\phi_i = B A \cos 30^{\circ} = 4 	imes (2.5 	imes 2) 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3} \ Wb$.However,standard interpretation of 'placed at $60^{\circ}$ to the field' often implies $	heta = 60^{\circ}$ as the angle between the normal and the field. Using $\phi = B A \cos 60^{\circ}$:$\phi_i = 4 	imes 5 	imes 0.5 = 10 \ Wb$.Final flux $\phi_f = 0 \ Wb$ (as the loop is removed from the field).Average induced emf $\varepsilon = -\frac{\Delta \phi}{\Delta t} = -\frac{\phi_f - \phi_i}{\Delta t} = -\frac{0 - 10}{10} = +1 \ V$.

$A$ rectangular loop of length $2.5 \ m$ and width $2 \ m$ is placed at $60^{\circ}$ to a magnetic field of $4 \ T$. The loop is removed from the field in $10 \ s$. The average emf induced in the loop during this time is

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