In an experiment designed to measure the magnetic field in the centre of a solen

Question

In an experiment designed to measure the magnetic field in the centre of a solenoid, the apparatus below was used. The loop is balanced when no current flows in the solenoid. When current flows in the solenoid and in the loop circuit, the loop experiences a downward force on the end inserted in the solenoid. To rebalance the loop, a weight of 20 timed 10^-4 N is required. (a) List two changes you could make to the circuit(s) that would change the direction of the force acting on the loop. (b) If the loop is 2 cm long, and a current of 2 A flows through the solenoid, calculate the magnetic force in the solenoid. (c) In measuring the loop, the end piece only was considered. Explain why the long sides of the loop were not considered. They are parallel to the field, therefore experience no force.

Explanation / Answer

Solution :- a) First thing we can do is that we can install the spring mounted weight to balance out the downward force. Secondly we can change the direction of current and then weight can be suitably added.

b) Length of loop = 2cm = 2/100 m = 0.02 m

Curent, I = 2A

Since B is not given so we will assume it as unity.

So magnetic force = BIL = 0.02* 2*1 = 0.04 N

c) The long side of the loop were not considered because they are actually parallel to the field and hence it will not experience any force. That is the reason it has not been considered.

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