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# Tutor profile: Aiswarya V.

Aiswarya V.
Passionate about teaching Maths, Physics, Programming and concepts in Electrical and Computer Engineering

## Questions

### Subject:Physics (Electricity and Magnetism)

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Question:

A straight wire of mass 300 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field $$B$$. What is the magnitude of the magnetic field?

Aiswarya V.

Given quantities are : mass $$m$$ = 300g = 0.3kg, length $$l$$ = 1.5m, current $$I$$ = 2A In order for the current carrying wire to be suspended mid air, the weight of the wire (downward force due to gravity) must be balanced by an upward Lorentz force due to the magnetic field. The Lorentz force is the force experienced by electric charge in motion due to a magnetic field. Taking acceleration due to gravity $$g$$ = 9.8$$m/s^{2}$$, weight of the wire = $$mg$$ Lorentz force due to magnetic field = $$IlB$$ Balancing the two forces out we get, $$IlB$$ = $$mg$$ $$B$$ = $$mg$$ / $$Il$$ $$B$$ = (0.3x9.8) / (2x1.5) $$B$$ = 0.98 T Therefore the magnitude of the magnetic field is 0.98 T.

### Subject:Linear Algebra

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Question:

Let A be a square matrix of order 3 x 3, then determinant $$|2A|$$ is equal to (A) $$2 | A |$$ (B) $$4 | A |$$ (C) $$6 | A |$$ (D) $$8 | A |$$

Aiswarya V.

Solution : (D) $$8|A|$$ Reason : Let matrix $$A$$ be given by $$A_{3,3} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$$. Then $$2A$$=$$\begin{pmatrix} 2a_{11} & 2a_{12} & 2a_{13} \\ 2a_{21} & 2a_{22} & 2a_{23} \\ 2a_{31} & 2a_{32} & 2a_{33} \end{pmatrix}$$, as every element in $$A$$ gets multiplied by the scalar quantity 2. Then $$|2A|$$ = $$2a_{11}$$($$2a_{22}$$x$$2a_{33}$$-$$2a_{23}$$x$$2a_{32}$$) - $$2a_{12}$$($$2a_{21}$$x$$2a_{33}$$ - $$2a_{23}$$x$$2a_{31}$$) + $$2a_{13}$$($$2a_{21}$$x$$2a_{32}$$ - $$2a_{22}$$x$$2a_{31}$$) =$$2a_{11}$$($$4a_{22}$$x$$a_{33}$$-$$4a_{23}$$x$$a_{32}$$) - $$2a_{12}$$($$4a_{21}$$x$$a_{33}$$ - $$4a_{23}$$x$$a_{31}$$) + $$2a_{13}$$($$4a_{21}$$x$$a_{32}$$ - $$4a_{22}$$x$$a_{31}$$) Taking 4 as common from out of the brackets we get $$|2A|$$ = $$2a_{11}$$x$$4$$($$a_{22}$$x$$a_{33}$$-$$a_{23}$$x$$a_{32}$$) - $$2a_{12}$$x$$4$$($$a_{21}$$x$$a_{33}$$ - $$a_{23}$$x$$a_{31}$$) + $$2a_{13}$$x$$4$$($$a_{21}$$x$$a_{32}$$ - $$a_{22}$$x$$a_{31}$$) = $$8a_{11}$$($$a_{22}$$x$$a_{33}$$-$$a_{23}$$x$$a_{32}$$) - $$8a_{12}$$($$a_{21}$$x$$a_{33}$$ - $$a_{23}$$x$$a_{31}$$) + $$8a_{13}$$($$a_{21}$$x$$a_{32}$$ - $$a_{22}$$x$$a_{31}$$) =$$8$$[$$a_{11}$$($$a_{22}$$x$$a_{33}$$-$$a_{23}$$x$$a_{32}$$) - $$a_{12}$$($$a_{21}$$x$$a_{33}$$ - $$a_{23}$$x$$a_{31}$$) + $$a_{13}$$($$a_{21}$$x$$a_{32}$$ - $$a_{22}$$x$$a_{31}$$)] =$$8|A|$$ Hence option (D) is correct. The above solution can be generalised as follows : For a scalar k and an $$nxn$$ matrix $$A$$ $$|kA|$$=$$k^{n}$$$$|A|$$.

### Subject:Calculus

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Question:

Discuss the continuity of the modulus function $$f(x)=|x|$$ at $$x=0$$.

Aiswarya V.

The modulus function $$f(x)=|x|$$ is defined as $$f(x)= -x$$, if $$x<0$$ and $$f(x)= x$$ , if $$x>=0$$. To know if a function is continuous at a point intuitively, we just need to check if approaching the point from any side of the graph gives the same value as the function itself at that point. This means that the left hand limit and right hand limit must be equal to the value of the function at $$x=0$$. We are taking both left hand and right hand limit at zero because the modulus function is defined differently on both sides of zero (as defined above) and so we need to make sure that it give the same value from both sides. The value of the function at $$x=0$$ is clearly defined : $$f(0)=0$$. The left hand limit is given by $$\lim\limits_{x \to 0^{-}} f(x) =$$ $$\lim\limits_{x \to 0^{-}} (-x) = 0$$. The right hand limit is given by $$\lim\limits_{x \to 0^{+}} f(x) =$$ $$\lim\limits_{x \to 0^{+}} (x) = 0$$. Since the left hand limit, right hand limit, and the value of function at $$x=0$$ coincide, the modulus function is continuous at $$x=0$$.

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