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1

### IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the xy-plane and a steady current I flows through the wire. The z-component of the magnetic field at the centre of the spiral is A
$${{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)$$
B
$${{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{{b + a} \over {b - a}}} \right)$$
C
$${{{\mu _0}NI} \over {2b}}\ln \left( {{b \over a}} \right)$$
D
$${{{\mu _0}NI} \over {2b}}\ln \left( {{{b + a} \over {b - a}}} \right)$$

## Explanation

Magnetic field at the centre of a circular loop of radius r and carrying a current $$I = {{{\mu _0}I} \over {2r}}$$. The direction of the field is along z-direction if the current is anticlockwise.

Consider a small element of width dr. The current through the element is

$$dI = {{total\,current\,in\,spiral} \over {total\,width\,of\,spiral}} \times width\,of\,element$$

$$= {{Idr} \over {(b - a)}}$$

$$\therefore$$ $$B = \int\limits_a^b {{{{\mu _0}NdI} \over {2r}} = \int\limits_a^b {{{{\mu _0}NI} \over {2(b - a)}}{{dr} \over r}} }$$

$$= {{{\mu _0}NI} \over {2(b - a)}}\int\limits_a^b {{{dr} \over r} = {{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)}$$

2

### IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature Tc(0). An interesting property of superconductors is that their critical temperature becomes smaller than Tc(0), if they are placed in a magnetic field, that is, the critical temperature Tc(B) is a function of the magnetic field strength B. The dependence of Tc(B) on B is shown in the figure. A superconductor has Tc(0) = 100 K. When a magnetic field of 7.5 T is applied, its Tc decreases to 75 K. For this material, one can definitely say that when

A
B = 5 T, Tc(B) = 80 K
B
B = 5 T, 75 K < Tc (B) < 100 K
C
B = 10 T, 75 K < Tc < 100 K
D
B = 10, Tc = 70 K

## Explanation

It is given that Tc(0) = 100 K and Tc(7.5) = 75 K. Since Tc(B) is a monotonically decreasing function of B, Tc(5) < Tc(0) and Tc(5) > Tc(7.5). Thus, 75 K < Tc(5) < 100 K.

3

### IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

Electrical resistance of certain materials, known as superconductors, changes abruptly from a non-zero value to zero as their temperature is lowered below a critical temperature Tc(0). An interesting property of superconductors is that their critical temperature becomes smaller than Tc(0), if they are placed in a magnetic field, that is, the critical temperature Tc(B) is a function of the magnetic field strength B. The dependence of Tc(B) on B is shown in the figure. In the graph below, the resistance R of a superconductor is shown as a friction of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1 which of the following graphs shows the correct variation of R with T in these fields?

A B C D ## Explanation

From the given figure, Tc(B) is a monotonically decreasing function of B. Thus, B2 > B1 implies Tc(B2) < Tc(B1). Hence resistance with B2 will become zero at lower temperature in comparison to B1.

4

### IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is A
IBL
B
$${{IBL} \over \pi }$$
C
$${{IBL} \over {2\pi }}$$
D
$${{IBL} \over {4\pi }}$$

## Explanation Consider an small element AB of length dl of the circle of radius R subtending an angle $$\theta$$ at the centre O.

If T is the tension in the wire, then force towards the centre will be equal to $$2T\sin \left( {{\theta \over 2}} \right)$$ which is balanced by outward magnetic force on the current carrying element $$( = IdlB)$$

$$2T\sin \left( {{\theta \over 2}} \right) = IdlB$$

For small angle $$\theta$$, $$\sin {\theta \over 2} \approx {\theta \over 2}$$

or, $$T = {{IBdl} \over \theta } = IBR$$ ($$\because$$ $$\theta = {{dl} \over R}$$)

$$= {{IBL} \over {2\pi }}$$ ($$\because$$ $$R = {L \over {2\pi }}$$)

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