A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth’s horizontal magnetic field is 24 μT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.

Given, Time period of oscillation, T1 = 0.10 sec


Horizontal component of Earth's magnetic field,



Downward current in the vertical wire, I =18A


Distance of wire from the magnet, d = 20cm =0.2m


When a current-carrying wire is placed near the magnet, the effective magnetic field gets changed. Now the net magnetic field can be obtained by subtracting the magnetic field due to the wire from Earth's magnetic field.



⇒






Time period of the coil (T) is given by



Let T1 and T2 be the time periods of the coil in the absence of the wire and in the presence the wire respectively. As time period is inversely proportional to magnetic field,



⇒



T2 = 0.076sec


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