krit.club logo

Magnetic Effects of Current and Magnetism - Torque on a Current Loop and Galvanometers

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A current-carrying loop placed in a uniform magnetic field B⃗\vec{B} experiences a torque, but the net force acting on it is zero. The torque tends to rotate the loop about its axis.

•

The Magnetic Dipole Moment m⃗\vec{m} of a current loop is defined as the product of the current II and the area vector A⃗\vec{A}. For NN turns, m⃗=NIA⃗\vec{m} = N I \vec{A}.

•

A Moving Coil Galvanometer (MCG) works on the principle that a current-carrying coil placed in a magnetic field experiences a torque, which is balanced by a restoring torque from a suspension wire or spring.

•

The magnetic field in a MCG is made radial by using concave pole pieces and a soft iron core. This ensures the torque is maximum and constant: τ=NIAB\tau = NIAB (since sin⁡θ=1\sin \theta = 1).

•

Current Sensitivity (SiS_i) is the deflection produced per unit current: Si=ϕIS_i = \frac{\phi}{I}. Voltage Sensitivity (SvS_v) is the deflection per unit voltage: Sv=ϕVS_v = \frac{\phi}{V}.

•

To convert a galvanometer into an ammeter, a low resistance called a 'shunt' (SS) is connected in parallel. To convert it into a voltmeter, a high resistance (RR) is connected in series.

📐Formulae

τ=NIABsin⁡θ\tau = NIAB \sin \theta

τ⃗=m⃗×B⃗\vec{\tau} = \vec{m} \times \vec{B}

m=NIAm = NIA

Equilibrium Condition for MCG: NIAB=kϕ\text{Equilibrium Condition for MCG: } NIAB = k \phi

Si=ϕI=NBAkS_i = \frac{\phi}{I} = \frac{NBA}{k}

Sv=ϕV=NBAkRS_v = \frac{\phi}{V} = \frac{NBA}{kR}

Shunt Resistance: S=(IgI−Ig)G\text{Shunt Resistance: } S = \left( \frac{I_g}{I - I_g} \right) G

Series Resistance: R=VIg−G\text{Series Resistance: } R = \frac{V}{I_g} - G

💡Examples

Problem 1:

A circular coil of 100100 turns and radius 5 cm5 \text{ cm} carries a current of 0.1 A0.1 \text{ A}. It is placed in a uniform magnetic field of 0.5 T0.5 \text{ T} such that the plane of the coil is parallel to the field. Calculate the torque acting on the coil.

Solution:

Given: N=100N = 100, r=0.05 mr = 0.05 \text{ m}, I=0.1 AI = 0.1 \text{ A}, B=0.5 TB = 0.5 \text{ T}. Since the plane is parallel to the field, the angle θ\theta between the area vector and BB is 90∘90^\circ. Area A=πr2=π(0.05)2=0.00785 m2A = \pi r^2 = \pi (0.05)^2 = 0.00785 \text{ m}^2. Torque τ=NIABsin⁡90∘=100×0.1×0.00785×0.5×1=0.03925 N m\tau = NIAB \sin 90^\circ = 100 \times 0.1 \times 0.00785 \times 0.5 \times 1 = 0.03925 \text{ N m}.

Explanation:

The torque is maximum when the plane of the loop is parallel to the magnetic field because the normal to the loop is perpendicular to the field lines.

Problem 2:

A galvanometer has a resistance of 15 Ω15 \text{ } \Omega and shows full scale deflection for a current of 4 mA4 \text{ mA}. How will you convert it into an ammeter of range 00 to 6 A6 \text{ A}?

Solution:

Given G=15 ΩG = 15 \text{ } \Omega, Ig=4 mA=4×10−3 AI_g = 4 \text{ mA} = 4 \times 10^{-3} \text{ A}, I=6 AI = 6 \text{ A}. Shunt resistance S=IgGI−IgS = \frac{I_g G}{I - I_g}. S=0.004×156−0.004=0.065.996≈0.01 ΩS = \frac{0.004 \times 15}{6 - 0.004} = \frac{0.06}{5.996} \approx 0.01 \text{ } \Omega.

Explanation:

To convert a galvanometer into an ammeter, a very small resistance (shunt) must be connected in parallel so that most of the current bypasses the sensitive galvanometer coil.

Torque on a Current Loop and Galvanometers Revision - Class 12 Physics ICSE