A circular coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.10T` normal to the place of the coil. If the current in the coil is `5.0A`, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) Given, `N=10^(29)m^-3`, `A=10^-5m^2`
Force on an electron of charge e, moving with drift velocity `v_d` in the magnetic field is given by
`F=Bev_d=Be(I)/(NeA) ( :'I=NeAv_d)`
`F=(BI)/(NA)=(0*10xx5*0)/(10^(29)xx10^-5)=5xx10^(-25)N`

Explanation

To solve the given problem step by step, we will address each part of the question separately.

### Given Data:
- Number of turns in the coil, \( N = 20 \)
- Radius of the coil, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Magnetic field strength, \( B = 0.10 \, \text{T} \)
- Current in the coil, \( I = 5.0 \, \text{A} \)

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