Answer:
The magnetic field along x axis is

The magnetic field along y axis is zero.
The magnetic field along z axis is

Explanation:
Given that,
Length of the current element 
Current in y direction = 5.40 A
Point P located at 
The distance is


We need to calculate the magnetic field
Using Biot-savart law

Put the value into the formula

We need to calculate the value of 



Put the value into the formula of magnetic field


Hence, The magnetic field along x axis is

The magnetic field along y axis is zero.
The magnetic field along z axis is

Answer:
Melt.
Explanation:
When rocks melt, they do so slowly and gradually because most rocks are made of several minerals, which all have different melting points; moreover, the physical and chemical relationships controlling the melting are complex. As a rock melts, for example, its volume changes. When enough rock is melted, the small globules of melt link up and soften the rock.
Under normal conditions, mantle rock like peridotite shouldn't melt in the Earth's upper mantle. However, by adding water you can lower the melting point of the rock. Alternatively, by decompressing the rock, you can bring it to a pressure where the melting point is lower. In both cases, basalt magma will form and considering it is hotter and less dense than the surrounding rock, it will percolate towards the surface and some of that erupts.
Answer:
Explanation:
Given
mass of disk 
diameter of disc 
Force applied 
Now this force will Produce a torque of magnitude



And Torque is given Product of moment of inertia and angular acceleration 

Moment of inertia for Disc 


