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Calcuting magnetic moment and current induced

2 votes

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.

asked May 4, 2017 in Physics Problems by Green (564 points)
edited May 15, 2018 by sammy gerbil
I believe typing the question atleast would make it easily searchable for the search engines and even this site itself ,  

Additionally if a link to the original question may be given.
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
Please can you show your calculation? Does your answer apply to case (A)?
@sammygerbil I tried to apply formula emf=rate of change of magnetic flux .but none of the options match.
Please show your calculation.

1 Answer

0 votes

Your answer is almost correct. Probably you made a mistake with the algebra.

What is misleading about this question is that answers in Column II can apply to more than one question in Column I, and more than answer can apply to the same question.

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.

So the answers are :

(A) = (s, t) (B) = (r, t) (C) = (t only), (D) = (r, t)

answered May 16, 2018 by sammy gerbil (27,556 points)
edited May 17, 2018 by sammy gerbil