# Calcuting magnetic moment and current induced

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Here is the question I am facing trouble with.

In this question I tried to find out the induced emf and then the current but the options did not match. My answer was $emf=a^2\alpha/2$ in different faces.

edited May 15, 2018
I believe typing the question atleast would make it easily searchable for the search engines and even this site itself ,

as sometimes the words of the question are crucial.
Yes I tried putting out the link of image in .jpg format.But I couldn't do that with pasteboard.do you have any other site which offers a direct link in .jpg format.
Use cdn link in pasteboard .
Or you can use imgur.com
@sammygerbil I tried to apply formula emf=rate of change of magnetic flux .but none of the options match.

The magnitude of the magnetic field $\vec{B}=\alpha t \hat{n}$ is the same in all 4 cases. Only the direction varies. Here $\hat{n}=(b_x\hat{i}+b_y\hat{j}+b_z\hat{k})$ is a unit vector - ie $b_x^2+b_y^2+b_z^2=1$.

Examine what happens with the $x$ component $b_x$.

The magnetic flux through each of the 2 square faces with normals in the $\hat{i}$ direction is $a^2 (\alpha b_x t)$. The rate of change of this flux is $a^2 \alpha b_x$. Applying Faraday's Law, an emf of $a^2 \alpha b_x$ is induced in each of the square loops surrounding these 2 faces. The current around each loop is $I_x=\frac{a^2 \alpha b_x}{4ar}=\frac{a\alpha b_x}{4r}$. The magnetic moment of the two loops together is $m_x=2I_x a^2 =\frac{a^3 \alpha}{2r}b_x$.

A similar calculation can be done for the $B_y, B_z$ components also. The magnitude of the total magnetic moment of the cube is then
$m=\sqrt{m_x^2+m_y^2+m_z^2}=\frac{a^3 \alpha}{2r} \sqrt{b_x^2+b_y^2+b_z^2}=\frac{a^3\alpha}{2r}$

The magnitude of the magnetic moment is the same in all 4 cases : it is independent of the direction of the magnetic field. Answer (t) matches questions (A), (B), (C) and (D).

Edge AB is common to 2 faces, with normals in the $+\hat{i}, +\hat{j}$ directions. If $b_x, b_y$ are +ve then the currents induced in the loops around these 2 faces are both anticlockwise - ie in the direction AB in both loops.

So the current in AB is $I_x+I_y=\frac{a\alpha}{4r}(b_x+b_y)$.

In questions (B) and (D) $b_x+b_y=0$ so the current in AB is zero - answer (r).

In question (A) $b_x+b_y=1$ so the current in AB is $\frac{a\alpha}{4r}$ - answer (s).

In question (C_ $b_x+b_y=-\frac{1}{\sqrt7}$ so the current in AB is $-\frac{a\alpha}{4r\sqrt7}$ - which does not correspond to any answer.