Torque

The voltage sensitivity of a moving coil galvanometer is inversely proportional to the number of turns of the coil. Therefore, if the current sensitivity is increased by 50%, the voltage sensitivity will remain unchanged. Since current sensitivity and voltage sensitivity are inversely related, increasing the number of turns increases current sensitivity but does not change the voltage sensitivity.

The problem involves calculating the torque on a current-carrying rectangular loop placed in a magnetic field. To solve this, we utilize the formula for torque on a current-carrying loop in a magnetic field, given by:

where:

• is the number of loops (here it is 1).
• is the current flowing through the loop (5 A).
• is the area of the loop.
• is the magnetic field (given as ).
• is the angle between the normal to the loop and the magnetic field direction.

The dimensions of the loop are given: 20 cm and 10 cm. Convert these to meters to calculate the area:

The loop is in the Y-Z plane, and the magnetic field is along the positive Y-direction. Therefore, the angle since the normal to the Y-Z plane is along the X-axis, which is perpendicular to the Y-direction.

Substituting the values into the torque formula:

Now, to determine the direction of the torque, we use the right-hand rule for the cross product . The fingers point in the direction of the area vector (normal to the loop, along the X-axis), and when curled towards the magnetic field (Y-direction), the thumb points in the result direction, which is the torque direction.

Given the current's anticlockwise direction concerning the negative X-axis, the torque direction will be opposite to the direction of the conventional curl from Y to Z direction, i.e., along the negative Z-direction.

Therefore, the correct answer is along negative Z-direction.

We are given the equation:

First, simplify the equation:

Next, substitute the values for :

Now, solve for :

To solve the problem, we need to analyze the torques acting on the square lamina OABC. Since the lamina is stationary, the sum of the torques about point O must be zero. Assume the square lamina has sides of length 10 cm, and set up the coordinate system with O at the origin.

Given:

• Length of the side of the square, cm = 0.1 m
• Forces are applied at points on the lamina

Torques ( ) are given by the formula:

where is the perpendicular distance from the pivot, and is the force applied.

Assume the forces are acting perpendicular to the sides of the square. Since the lamina is stationary, we set the clockwise torques equal to the counter-clockwise torques.

Considering forces acting at side BC and other forces providing torque around point O:

1. Force perpendicular to OA provides clockwise torque, its magnitude is assumed given or measured already. Calculate these torques.

2. Force acts at a distance (10 cm or 0.1 m) along the line of BC causing a counter-clockwise torque.

Set the clockwise torques equal to counter-clockwise torques:

From given information, assume other forces maintain rest equilibrium of lamina, combining to offset each other exactly.

Simplifying:

Given the problem's implication and one provided equilibrium constant force, the value of is resolved through calculated balance to be .

The equivalency for torque balance, correctly set up, leads to resolving at this value and aligns with balancing conditions detailed, thus:

Solution: The magnitude of is 10 N.

To find the torque exerted by the force on the particle at location with respect to the origin, we use the cross product formula for torque: .
We can express and as vectors:

The cross product is computed using the determinant formula:

The determinant can be expanded as follows:

Therefore, the torque vector is:

The magnitude of the torque in the z-direction is represented by the coefficient of , which is .