Chapter 25, Problem 44A

To determine

To Calculate: The amount of current the galvanometer will indicate.

Answer to Problem 44A

  $2.3\text{μA}$

Explanation of Solution

Given:

Length of the wire: $10\text{cm}$

Velocity of wire: $1.0\text{m/s}$

Magnetic Field: $2.0\times {10}^{-2}\text{T}$

Resistance of galvanometer: $875\Omega$

Angle between the motion of the wire and magnetic field: $\theta =90°$

Formula Used:

  $EMF=BLv\left(\sin \theta \right)$

Where,

  $EMF$ : Electromotive Force

  $B$ : Magnetic Field

  $L$ : Length

  $v$ : Velocity

  $I=\frac{EMF}{R}$

Where,

  $I$ : Current

  $R$ : Resistance

  $10\text{cm = 0}\text{.100}\text{m}$

Calculation:

In the first step, calculate the EMF of the galvanometer,

  $EMF=BLv\left(\sin \theta \right)$

  $=\left(\text{2}{\text{.0×10}}^{\text{-2}}\text{T}\right)\times \left(\text{0}\text{.100}\text{m}\right)\times \left(\text{1}\text{.0}\text{m/s}\right)\left(\sin 90°\right)$

  $=$ $2.0\times {10}^{-3}\text{V}$

Now to calculate the current use the formula of current,

  $I=\frac{EMF}{R}$

Substitute all the given values in the above formula

  $I$ $=$ $\frac{\text{2}{\text{.0×10}}^{\text{-3}}\text{V}}{\text{875}\text{Ω}}$

  $I$ $=$ $2.3\text{μA}$

Conclusion:

Hence,the amount of current the galvanometer will indicate is $2.3\text{μA}\text{.}$