Magnetic Force & Torque

1 Mark Questions

1. Using the concept of force between two infinitely long parallel current carrying conductors,

define one ampere of current. [All India 2014]

Ans.

Force between two straight parallel current carrying conductors

<math display="block">F = \frac{\mu_0}{4\pi} \frac{2I_1I_2}{r}</math>

Let <math>I_1 = I_2 = 1 \text{ A}, r = 1 \text{ m}</math>, then

<math>F = 10^{-7} \times \frac{2(1)(1)}{(1)} = 2 \times 10^{-7}</math> (2) One ampere One ampere is that value of current which flows through two straight, parallel infinitely long current carrying conductors placed in air or vacuum at a distance of 1 m and they experience a force of attractive or repulsive nature of magnitude <math>2 \times 10^{-7}</math> N/m on their unit length. (1)

2.Is the steady electric current the only source of magnetic field? Justify your answer.[Delhi 2013 C]

Ans. Yes, the net magnetic force acting on a wire carrying a steady (constant) electric current! in an external magnetic field Band is given F=IdIB

1. Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant? [All India 2008]

Ans.Low torsional constant facilitates greater deflection (allpha) in coil for given value of current and hence, sensitivity of galvanometer increases,

1. Write two factors by which voltage sensitivity of a galvanometer can be increased.[Foreign 2008]

Ans. Voltage sensitivity of galvanometer can be increased by

1. increasing the magnetic field (ii) decreasing the value of torsional constant,
2. Write two factors by which current sensitivity of a moving coil galvanometer can be increased. [Foreign 2008]

Ans. Current sensitivity of galvanometer can be increased by

1. increasing the number of turns of coil.
2. increasing the magnetic field.

2 Marks Questions

6.A rectangular coil of sider I and b carrying a current I is subjected to a uniform magnetic field B acting perpendicular to its plane. Obtain the expression for the torque acting on it. [Delhi 2014 C]

Ans.

Equivalent magnetic moment of the coil

<math>\mathbf{m} = IA\hat{\mathbf{n}}</math>

• ٠. <math>\mathbf{m} = Ilb \,\hat{\mathbf{n}}</math>

Wher, <math>\hat{\mathbf{n}} = \text{unit vector } \perp \text{ to the plane of the}</math> coil.

<math>\therefore</math> Torque = <math>\mathbf{m} \times \mathbf{B}</math> <math>= Ilb \hat{\mathbf{n}} \times \mathbf{B}</math> <math>=0</math> (1) As <math>\hat{\mathbf{n}}</math> and <math>\mathbf{B}</math> are parallel or antiparallel to each other.

7.A coil of N turns and radius R carries a current /. It is unwound and rewound to make a square coil of side a having same number of turns N. Keeping the current I same, find the ratio of the magnetic moments of the square coil and the Circular COil. [Delhi 2013 C]

Ans.

Ratio of the magnetic moments

<math display="block">\frac{M_s}{M_c} = \frac{2INA_s}{INA_c}</math>

<math>=\frac{2\left(\frac{R}{2}\right)^2}{(R)^2}=\frac{1}{2}</math>

8.A steady current I1 flows through a long straight wire. Another wire carrying steady current I₂ in the same direction is kept close and parallel to the first wire. Show with the help of a diagram, how the magnetic field due to the current I_r exert a magnetic force on the second wire. Deduce the expression for this force. [All India 2011; Foreign 2008] Ans.

In these types of questions, we are calculating force on a wire in the field produced by the other current carrying wire.

Let two infinitely long straight current carrying conductor carry currents <math>I_1</math> and <math>I_2</math> in the same direction. Magnetic field <math>B_1</math> due to first wire on seconds, i.e. (1/2)

<math display="block">B_1 = \frac{\mu_0}{4\pi} \frac{2I_1}{r}</math>

...(i)

(1/2)

The magnetic field is perpendicular to the plane of paper and directed inwards i.e. (X) type. × × × × 12 × × × × × × × × / · × × × ×