In a moving coil galvanometer, current (I) passing through galvanometer is directly proportional to its deflection (&ta;), i.e.
I = G&ta;
Where G = \(\frac {k}{NAB}\) ➔ Galvanometer constant
N = number of turns in coil; A = area of coil; B = strength of magnetic field; k = torsional constant of spring that means restoring torque per unit twist.

Field along axis of coil → B =\( \frac {\mu_o IR^2}{2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] } \)
At coil of coil → B₁ = \( \frac {\mu_o I}{2R}\)
\(\frac {B_1}{B} = \frac {\mu_o I}{2R} \, \times \, \frac {2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] }{\mu_o IR^2} = \frac {2 \big ( (R^2 \, + \, x^2)^{\frac {3}{2}} \big ] }{R^3}\)
B1 = \(\frac {48 [4^2 + 7^2]^{\frac {3}{2}}}{4^3}\)
B1 = \(\frac {48 \times (65)^{\frac {3}{2}}}{4^3}\)
B1 = 131.01 × 3
B1 = 393.03 = 393 μT

Cyclotron frequency = \(\frac {qB}{2 \pi m}\).
B = \(\frac {(2 \pi mf_c)}{(q)}\)
B = \(\frac {(2 \, \times 3.142 \, \times 3.3 \, \times \, 10^{-27} \, \times \, 12 \, \times \, 10^6)}{(1.6 \, \times \, 10^{-19})}\)
B = 1.56 T

Magnetic permeability = Magnetic flux density × [Magnetic field strength]^-1.
μ = [M L⁰ T^-2 A^-1] × [M⁰ L^-1 T⁰ A¹]^-1
μ = [M L T^-2 A^-2].

When a charge moves parallel or antiparallel to direction of magnetic field, it experiences a minimum (zero) force.
When &ta; = 0^o or 180^o,
F_m = qvBsin&ta; = qvB (0) = 0.

Yes, when infinitely long parallel wires carry equal currents in same direction, magnetic field at midpoint in between two wires is zero. This is because, when wires carry equal currents in same direction, magnetic field created at midpoint will cancel out each or and thus, field is zero.

F = qvB.
For α-particle, q = 2e, F_α = 2evB
For β-particle, q = e, F_β = evB.
Thus, α-particle will experience maximum force.

The product of pole strength of eir magnetic pole and magnetic length of magnetic dipole is called its magnetic dipole moment. The SI unit of magnetic dipole moment is Am².

Current sensitivity is defined as deflection produced in galvanometer, when unit current flows through it.
I_S = \(\frac {NAB}{k}\) ??????1
The unit of current sensitivity is rad A^-1 or div A^-1.
From 1, we can understand that current sensitivity is expressed as exact reverse of galvanometer constant (G = \(\frac {k}{NAB}\)).

Magnetic field at center of a circular arc of radius R, carries a current I and makes an angle &ta; at center is given by
B = \(\frac {\mu_o I \ta}{4 \pi R}\)
In this case &ta; = \(\frac {\pi }{4}\)
B = \(\frac {\mu_o I (\frac {\pi }{4})}{4 \pi R}\)
B = \(\frac {\mu_o I}{16 R}\)