Magnetic Dipole & Magnetic Field Lines

2 Marks Questions

1.Draw the magnetic field lines due to a current passing through a long solenoid. Use Ampere's circuital law, to obtain the expression for the magnetic field due to the current I in a long solenoid having n number of turns per unit length. [Delhi 2014c] (B) • В Applying Ampere's circuital law for the rectangular loop abcd,

<math display="block">\oint \mathbf{B} \cdot d\mathbf{I} = \mu_0 I</math> <math display="block">Bh = \mu_0 I(nh)</math> <math>B = \mu_0 nI</math> (1) a ºB₁ b B₂

<math>I_a</math> <math>I_b</math>

Q <math>F_2</math> Ρ а Let a and b be two long straight parallel conductors. <math>I_a</math> and <math>I_b</math> are the current flowing through them and separated by a distance d. Magnetic field induction at a point P on a conductor b due to current I passing through a is

2.(i) Two long straight parallel conductors a and b carrying steady currents Ia and Ib respectively are separated by a distance d. Write the magnitude and direction, what is the nature and magnitude of the force between the two conductors?

1. Show with the help of a diagram, how the force between the two conductors would

change when the currents in them flow in the opposite directions. [Foreign 2014]

<math display="block">B_1 = \frac{\mu_0 2I_a}{4\pi d}</math>

Now, unit length of b will experience a force as

<math display="block">F_2 = B_1 I_b \times 1 = B_1 I_b</math> <math display="block">F_2 = \frac{\mu_0}{4\pi} \frac{2I_a I_b}{d}</math> ٠. Conductor a also experiences the same amount of force directed towards b. Hence, a and b attract each other. (1)

(ii)

<math>I_b</math>

Q P <math>F_1</math>

Now, let the direction of current in conductor b be reversed. The magnetic field B₂ at point P due to current Ia flowing through a will be downwards. Similarly, the magnetic field B₁ at point Q due to current Ib passing through b will also be downward as shown. The force on a will be, therefore towards the left. Also, the force on b will be towards the right. Hence, the two conductors will repel each other as shown.

3.A circular coil of N turns and radius R carries a current L It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. [All India 2012]

<math>\langle \rangle</math> The length of wire will be same in two cases as the same coil is unwound and rewound.

Length of the wire is same

<math display="block">\therefore N_1 \times (2\pi R) = N_2 \times 2\pi \left(\frac{R}{2}\right)</math>

<math>[N_1 \text{ and } N_2 = \text{number of turns in two coils}]</math>

<math>N_2 = 2N_1</math>

(1/2)

Now, the ratio of magnetic moments is given by

<math display="block">\frac{M_1}{M_2} = \frac{N_1 I A_1}{N_2 I A_2} = \frac{N_1 \times \pi R_1^2}{N_2 \times \pi R_2^2}</math>

(1/2)

<math>\frac{M_1}{M_2} = \left(\frac{N_1}{2N_1}\right) \times \left(\frac{R}{R/2}\right)^2 = \frac{1}{2} \times 4 = 2</math>

(1/2)

<math>M_1: M_2 = 2:1</math>

(1/2)

4.A circular coil of N turns and diameter d carries a current I. It is unwound and rewound to make another coil of diameter 2d, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. [All India 2012]